Arithmetic Progressions

Class 10 NCERT

NCERT

1   In which of the following situations, does the list of numbers involved make as arithmetic progression and why? $\\$ (i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km. $\\$ (ii) The amount of air present in a cylinder when a vacuum pump removes $ \frac{1}{4}$ of the air remaining in the cylinder at a time. $\\$ (iii) The cost of digging a well after every metre of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent metre. (iv)The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8% per annum.

Solution :

In other words, after every stroke, only $ 1-\frac{1}{4}=\frac{3}{4} \ $part of air will remain.$\\$ Therefore, volumes will be V, $ \frac{3V}{4},(\frac{3V}{4})^2,\frac{3V}{4})^3 ....$ Clearly, it can be observed that the adjacent terms of this series do not have the same difference between them. Therefore, this is not an A.P.$\\$ (iii) Cost of digging for first metre = 150$\\$ Cost of digging for first 2 metres = 150 + 50 = 200$\\$ Cost of digging for first 3 metres = 200 + 50 = 250$\\$ Cost of digging for first 4 metres = 250 + 50 = 300$\\$ Clearly, 150, 200, 250, 300 ... forms an A.P. because every term is 50 more than the preceding term.$\\$ (iv) We know that if Rs P is deposited at r% compound interest per annum for n years, our money will be $P(1+\frac{r}{100})^n $ after n years. Therefore, after every year, our money will be

(ii) Let the initial volume of air in a cylinder be V lit. In each stroke, the vacuum pump removes $\frac{1}{4}$ of air remaining in the cylinder at a time.

$10000(1+\frac{8}{10}),10000(1+\frac{8}{10})^2,10000(1+\frac{8}{10})^3,10000(1+\frac{8}{10})^4......\\$ Clearly, adjacent terms of this series do not have the same difference between them. Therefore, this is not an A.P.

(i) It can be observed that Taxi fare for 1st km = 15 $\\$ Taxi fare for first 2 km = 15 + 8 = 23 $\\$ Taxi fare for first 3 km = 23 + 8 = 31$\\$ Taxi fare for first 4 km = 31 + 8 = 39$\\$ Clearly 15, 23, 31, 39 ... forms an A.P. because every term is 8 more than the preceding term.

2   Write first four terms of the A.P. when the first term a and the common difference d are given as follows $\\$ (i) a = 10, d = 10 $\\$ (ii) a = -2, d = 0 $\\$ (iii) a = 4, d = -3 $\\$ (iv) a = -1 d = 0.5 $\\$ (v) a = -1.25, d = -0.25 $\\$

Solution :

(iii) a = 4, d = -3 Let the series be $a_1, a_2, a_3, a_4 ...\\ a_1 = a = 4\\ a_2 = a_1 + d = 4 - 3 = 1\\ a_3 = a_2 + d = 1 - 3 = -2\\ a_4 = a_3 + d = - 2 - 3 = -5\\$ Therefore, the series will be 4, 1, -2 -5 ...$\\$ First four terms of this A.P. will be 4, 1, -2 and -5.

(v) a = -1.25, d = -0.25 Let the series be $a_1, a_2, a_3, a_4 ...\\ a_1 =a=-1.25\\ a_2 = a_1 + d = -1.25 - 0.25 = -1.50\\ a_3 = a_2 + d = - 1.50 - 0.25 = -1.75\\ a_4 = a_3 + d = - 1.75 - 0.25 = -2.00\\$ Clearly, the series will be 1.25, -1.50, -1.75, -2.00 ........$\\$ First four terms of this A.P. will be -1.25, -1.50, -1.75 and -2.00.

(ii) a = -2, d = 0 Let the series be $a_1, a_2, a_3, a_4 ...\\ a_1 = a = -2\\ a_2 = a_1 + d = - 2 + 0 = -2\\ a_3 = a_2 + d = - 2 + 0 = -2\\ a_4 = a_3 + d = - 2 + 0 = -2\\$ Therefore, the series will be -2, -2, -2, -2 ...$\\$ First four terms of this A.P. will be -2, -2, -2 and -2.

a = -1, d = 0.5 Let the series be $ a_1, a_2, a_3, a_4 ...\\ a_1=a=-1\\ a_2=a_1+d=-1+\frac{1}{2}=-\frac{1}{2}\\ a_3=a_2+d=-\frac{1}{2}+\frac{1}{2}=0\\ a_4=a_3+d=0+\frac{1}{2}=\frac{1}{2}\\$ Clearly, the series will be $-1 , -\frac{1}{2},0,\frac{1}{2} .......$ First four terms of this A.P. will be $-1 , -\frac{1}{2},0,\frac{1}{2} .......$

i) a = 10, d = 10 $\\$ Let the series be $ a_1, a_2, a_3, a_4, a_5 ... \\ a_1 = a = 10 \\ a_2 = a_1 + d = 10 + 10 = 20\\ a_3 = a_2 + d = 20 + 10 = 30\\ a_4 = a_3 + d = 30 + 10 = 40 \\ a_5 = a_4 + d = 40 + 10 = 50 \\$ Therefore, the series will be 10, 20, 30, 40, 50 ... First four terms of this A.P. will be 10, 20, 30, and 40.

3   For the following A.P.s, write the first term and the common difference.$\\$ (i) 3, 1, - 1, - 3 ... $\\$(ii) - 5, -1, 3, 7 ...$\\$ (iii) $\frac{1}{3},\frac{5}{3},\frac{9}{3},\frac{13}{3}$ $\\$ (iv) 0.6, 1.7, 2.8, 3.9 ...

Solution :

(iv) 0.6, 1.7, 2.8, 3.9 ... $\\$ Here, first term, a = 0.6$\\$ Common difference, d = Second term - First term = 1.7 - 0.6 = 1.1

$\frac{1}{3},\frac{5}{3},\frac{9}{3},\frac{13}{3}$ $\\$ Here, first term, $\frac{1}{3}$ Common difference, d = Second term - First term =$\\$ $\frac{5}{3}-\frac{1}{3}=\frac{4}{3}$

(i) 3, 1, -1, -3 ...$\\$ Here, first term, a = 3$\\$ Common difference, d = Second term - First term = 1 - 3 = -2$\\$ (ii) -5, -1, 3, 7 ... $\\$ Here, first term, a = -5 Common difference, d = Second term - First term = (-1) -(-5) = - 1 + 5 = 4

4   For the following $A.P.$s, write the first term and the common difference. $\\$ $(i) 3, 1, - 1, - 3 \dots \\ (ii) - 5, - 1, 3, 7 \dots \\ (iii) 1 3 , 5 3 , 9 3 , 13 3 \dots \\ (iv) 0.6, 1.7, 2.8, 3.9 \dots $

Solution :

(i) The given $AP$ is $3, 1, -1, -3 \dots $$\\$ Here, first term, $a = 3$$\\$ Common difference, d = Second term - First term $\\$ $= 1 - 3\\ = -2 $ (ii) The given $AP$ is $ -5, -1, 3, 7 \dots $$\\$ Here, first term,$ a = -5$$\\$ Common difference, d = Second term - First term $\\$ $= (-1) -(-5) \\ = -1 + 5 \\ = 4 $$\\$ (iii) The given $AP$ is $\dfrac{1}{ 3} , \dfrac{5}{ 3} , \dfrac{9}{ 3} , \dfrac{13}{ 3} \dots $$\\$ Here, first term, $a = \dfrac{1}{ 3}$$\\$ Common difference, d = Second term - First term $=\dfrac{ 5}{ 3} - \dfrac{1}{ 3} = \dfrac{4}{ 3}$$\\$ (iv) The given $AP$ is $0.6, 1.7, 2.8, 3.9 \dots$$\\$ Here, first term, $a = 0.6$$\\$ Common difference, d = Second term - First term $= 1.7 - 0.6\\ = 1.1 $

5   Which of the following are $AP$s ? If they form an $AP$, find the common difference d and write three more terms.$\\$ $( i) 2, 4,8,16\dots \\ ( ii)2, 5 2 ,3, 7 2 ,\dots \\ ( iii ) 1.2, 3.2, 5.2, 7.2 \dots \\ ( iv) 10, 6, 2,2\dots \\ ( v)3,3 2,3 2 2,3 3 2 \dots.\\ ( vi)0.2,0.22,0.222,0.2222 \dots \\ ( vii )0, 4, 8, 12\dots \\ (viii)-\dfrac{1}{2},-\dfrac{1}{2},-\dfrac{1}{2},-\dfrac{1}{2}\dots $

Solution :

(iv) The given Series is $-10, - 6, - 2, 2 \dots $$\\$ First Term, $a_1= -10$$\\$ Second Term,$a_2 = -6$$\\$ Third Term, $a_3 = -2$$\\$ Fourth Term,$a_4 = -2$$\\$ Common difference $d,= a_2 - a_1 = (-6) - (-10) = 4 $$\\$ $= a_3 -a_2 = (-2) - (-6) = 4 \\ = a_4 - a_3 = (2) - (-2) = 4 $$\\$ As ,$a_2 - a_1 = a_3 - a_2 = a_4 -a_3$ they form an $AP$$\\$ Fifth Term, $a_5 = 2 + 4 = 6$$\\$ Sixth Term, $a_6 = 6 + 4 = 10$$\\$ Seventh Term, $a_7 = 10 + 4 = 14 $$\\$

(viii) To given Series is $-1 /2, -1 /2, -1 /2, -1 /2 \dots $$\\$ First Term, $a_1= -1/2$$\\$ Second Term,$ a_2 = -1/2$$\\$ Third Term, $a_3 = -1/2$$\\$ Fourth Term, $a_4 = -1/2 $$\\$ Common difference $d,= a_2 - a_1 = (-1 /2) - (-1 /2) = 0$$\\$ $ = a_3 -a_2 = (-1 /2) - (-1 /2) = 0\\ = a_4 -a_3 = (-1 /2) - (-1 /2) = 0 $$\\$ As ,$a_2 - a_1 = a_3 - a_2 = a_4 -a_3$ they form the $AP$$\\$ Series Fifth Term, $a_5 = (-1 /2) - 0 = -1 /2$$\\$ Sixth Term, $a_6 = (-1 /2) - 0 = -1 /2$$\\$ Seventh Term, $a_7 (-1 /2) - 0 = -1 /2 $$\\$ (ix) The gven Series is $1 , 3, 9, 27 \dots $$\\$ First Term, $a_1= 1 $$\\$ Second Term, $a_2 = 3$$\\$ Third Term, $a_3 = 9$$\\$ Fourth Term, $a_4 = 27 $$\\$ Common difference $d,= a_2 - a_1 = = 3 - 1 = 2 $$\\$ $= a_3 - a_2 = 9 - 3 = 6 = a_4 - a_3 = 27 - 9 = 1 8 $$\\$ As ,$a_2 - a_1 \neq a_3 - a_2$$\\$ Hence the given series does not form AP (x) The given Series is $a, 2a, 3a, 4a $$\\$ First Term, $a_1= a$$\\$ Second Term, $a_2 = 2a $$\\$ Third Term, $a_3 = 3a$$\\$ Fourth Term,$ a_4 = 4a$$\\$ Common difference $d,= a_2 - a_1 = 2a- a= a $$\\$ $= a_3 - a_2 = 3a-2a = a \\ = a_4 - a_3 = 4a-3a =a $$\\$ As ,$a_2 - a_1 = a_3 - a_2 = a4 - a3$ they form the $AP$$\\$ Series Fifth Term, $a_5 = 4a+ a =5a$$\\$ Sixth Term, $a_6 = 5a+a =6a$$\\$ Seventh Term, $a_7 = 6a+a =7a $$\\$

(iii) The given Series is $-1.2, - 3.2, -5.2, -7.2 \dots $$\\$ First Term, $a_1= 2$$\\$ Second Term,$ a_2 = 4$$\\$ Third Term, $a_3 = 8$$\\$ Fourth Term, $a_4 = -7.2$$\\$ Fifth Term, $a_5 = ?$$\\$ Sixth Term, $a_6 =?$$\\$ Seventh Term,$ a_7 = ? $$\\$ Common difference $d,= a_2 - a_1 = ( -3.2) - ( -1 .2) = -2$$\\$ $ = a_3 - a_2 = ( -5.2) - ( -3.2) = -2\\ = a_4 - a_3 = ( -7.2) - ( -5.2) = -2 $$\\$ As ,$a_2 - a_1 = a_3 - a_2 = a_4 - a_3$ they form an $ AP$ $\\$ Fifth Term, $a_5 = - 7.2 - 2 = - 9.2$$\\$ Sixth Term, $a_6 = - 9.2 - 2 = - 11 .2$$\\$ Seventh Term, $a_7 = - 11 .2 - 2 = - 1 3.2 $$\\$ (iv) The given Series is $-10, - 6, - 2, 2 \dots $$\\$ First Term, $a_1= -10$$\\$ Second Term,$a_2 = -6$$\\$ Third Term, $a_3 = -2$$\\$ Fourth Term,$a_4 = -2$$\\$ Common difference $d,= a_2 - a_1 = (-6) - (-10) = 4 $$\\$ $= a_3 -a_2 = (-2) - (-6) = 4 \\ = a_4 - a_3 = (2) - (-2) = 4 $$\\$ As ,$a_2 - a_1 = a_3 - a_2 = a_4 -a_3$ they form an $AP$$\\$ Fifth Term, $a_5 = 2 + 4 = 6$$\\$ Sixth Term, $a_6 = 6 + 4 = 10$$\\$ Seventh Term, $a_7 = 10 + 4 = 14 $$\\$

(v) The given Series is $3, 3 + \sqrt{2}, 3 + 2\sqrt{2}, 3 + 3\sqrt{2}$$\\$ First Term, $a_1= 3$$\\$ Second Term, $a_2 = 3 + \sqrt{2}$$\\$ Third Term,$ a_3 = 3 + 2\sqrt{2}$$\\$ Fourth Term, $a_4 = 3 + 3\sqrt{2}$$\\$ Common difference $d,= a_2 - a_1 = 3 + \sqrt{2} - 3 = \sqrt{2}$$\\$ $= a_3 - a_2 = (3 + 2\sqrt{2}) - (3 + \sqrt{2}) = \sqrt{2}\\ = a_4 - a_3 = (3 + 3\sqrt{2}) - (3 + 2\sqrt{2}) = \sqrt{2}$$\\$ As ,$a_2 - a_1 = a_3 - a_2 = a_4 - a_3 $ they form an $AP$$\\$ Fifth Term, $a_5 = (3 + \sqrt{2}) + \sqrt{2} = 3 + 4\sqrt{2}$$\\$ Sixth Term, $a_6 = (3 + 4\sqrt{2}) + \sqrt{2} = 3 + 5\sqrt{2} $$\\$ Seventh Term, $a_7 = (3 + 5\sqrt{2}) + \sqrt{2} = 3 + 6\sqrt{2}$$\\$ (vi) Te given Series is $ 0.2, 0.22, 0.222, 0.2222 \dots $$\\$ First Term, $a_1= 0.2$$\\$ Second Term, $a_2 = 0.22 $$\\$ Third Term,$ a_3 = 0.222$$\\$ Fourth Term, $a_4 = 0.2222$$\\$ Common difference $d,= a_2 - a_1 = 0.22 - 0.2 = 0.02 $$\\$ $= a_3 -a_2 = 0.222 - 0.22 = 0.002\\ a_4 - a_3 = 0.2222 - 0.222 = 0.0002 $$\\$ As ,$a_2 - a_1 \neq a_3 - a_2 \neq a_4 - a_3$$\\$ Hence the given series does not form an $AP$$\\$ (vii) To given Series is $0, -4, -8, -12 \dots $$\\$ First Term, $a_1= 0$$\\$ Second Term,$ a_2 = -4 $$\\$ Third Term, $a_3 = -8$$\\$ Fourth Term,$a_4 = -12$$\\$ Common difference $d,= a_2 - a_1 = (-4) - 0 = -4 $$\\$ $= a_3 - a_2 = (-8) - (-4) = -4 \\ = a_4 - a_3 = (-12) - (-8) = -4 $$\\$ As ,$a_2 - a_1 = a_3 - a_2 = a_4 - a_3$ they form an $AP$$\\$ Fifth Term, $a_5 = -12 - 4 = -16 $$\\$ Sixth Term, $a_6 = -16 - 4 = -20 $$\\$ Seventh Term,$ a_7 = -20 - 4 = -24 $$\\$

(i) The given Series is $2, 4,8 ,16\dots $$\\$ First Term,$ a_1= 2$$\\$ Second Term,$ a_2 = 4 $$\\$ Third Term,$ a3 = 8$$\\$ Common difference d,=$ a_2 - a_1 = 4 -2 =2$$\\$ $= a_3 - a_2 = 8 - 4 =4 $$\\$ As ,$a_2 - a_1 \neq a_3 – a_2$$\\$ Hence the given series does not form an $AP$$\\$ (ii) The given Series is $2, 5/2, 3, 7/2 \dots $$\\$ First Term, $a_1= 2$$\\$ Second Term, $a_2 = 5/2$$\\$ Third Term,$ a_3 = 3$$\\$ Fourth Term, $a_4 = 7/2$$\\$ Fifth Term, $a_5 = ? $$\\$ Sixth Term, $a_6 =?$$\\$ Seventh Term, $a_7 = ?$$\\$ Common difference $d,= a_2 - a_1 = 5/2-2 =1/2 \\ = a_3 - a_2 = 3- 5/4 =1/2$$\\$ As $a_2 - a_1 = a_3 - a_2$ , given series form an $AP$.$\\$ Fifth Term, $a_5 = 7/2 +1/2 =$$\\$ Sixth Term,$ a_6 = 4 + 1 /2 = 9/2$$ \\$ Seventh Term, $a_7 = 9/2 + 1 /2 = 5$$\\$

6   Fill in the blanks in the following table, given that $a$ is the first term, $d$ the common difference and $a_n$ the $n^{th}$ term of the $A.P.$ $\\$ $\begin{array}{|c|c|} \hline & a & b & n & a_n \\ \hline I &7 & 3 & 8 &... \\ \hline II &-18 & ... & 10 & 0 \\ \hline III &... & -3 & 18 &-5 \\ \hline IV &-18.9 & 2.5 & ... &3.6 \\ \hline V &3.5 & 0 & 105 &... \\ \hline \hline \end{array}$

Solution :

III. Given, $\\$ $\bullet $ Common Difference, $d = -3,$$\\$ $\bullet $ Number of terms,$ n = 18,$$\\$ $\bullet $ Nth term of the AP Series,$a_n = 5 $$\\$ $\bullet $ First Term,$ a=? $$\\$ $\bullet $ We know that the nth term of an A.P. Series,$\\$ $ a_n = a + (n -1) d -5\\ = a + (18 - 1) (-3)$(By Substituting) $-5 = a + (17) (-3) -5 \\ = a - 51\\ a - 51 = -5 \\ a = 51 - 5$$\\$ Hence, $a = 46$$\\$ IV. Given, $\\$ $\bullet $ First Term, $a = -18.9, $$\\$ $\bullet $ Common Difference, $d = 2.5,$$\\$ $\bullet $ Nth term of the AP Series, $\\$ $ a_n = 3.6, $$\\$ $\bullet $ Number of terms, $n = ?$$\\$ We know that the nth term of an A.P. Series,$\\$ $ a_n = a + (n -1) d\\ 3.6 = - 18.9 + (n - 1) 2.5$$\ \ \ $ (By Substituting) $\\$ $3.6 + 18.9 = (n - 1) 2.5 \\ 22.5 = (n - 1) 2.5 \\ (n-1)=\dfrac{22.5}{2.5}$ $\ \ \ $ (By transposing)$\\$ $ n-1=9 \\ n = 10$ Hence, $n = 10 $$\\$ V. Given, $\\$ $\bullet $ First Term,$ a = 3.5,$ $\\$ $\bullet $ Common Difference, $d = 0,$ $\\$ $\bullet $ Number of terms, $n = 105,$ $\\$ $\bullet $ Nth term of AP Series, $\\$ $\bullet $$ a_n = ?$ $\\$ We know that the $ n^{th }$ term of an A.P. Series,$\\$ $ a_n = a + (n - 1) d \\ a_n = 3.5 + (105 - 1) 0 $ (By Substituting)$\\$ $ a_n = 3.5 + 104 × 0\\ a_n = 3.5 $

II. Given, $\\$ $\bullet $ First Term,$a = -18,$$\\$ $\bullet $ nth term of an AP Series, $a_n = 0, $$\\$ $\bullet $ Number of terms, $n = 10,$$\\$ $\bullet $ Common Difference,$ d = ?$$\\$ We know that the nth term of an A.P. Series, $\\$ $a_n = a + (n - 1) d \\ 0 = - 18 + (10 - 1) d$ (By Substituting) $18 = 9d\\ 9d = 18 \\ d = \frac{18}{ 9} = 2$ (By Transposing) Hence, common difference, $d = 2$

Given, $\\$ $\bullet $ First Term, $a = 7,$ $\\$ $\bullet $ Common Difference, $d = 3,$$\\$ $\bullet $ Number of Terms,$ n = 8,$$\\$ $\bullet n^{th}$ term of an AP Series, $a_n = ?$$\\$ We know that the $n^{th}$ term of an A.P. Series, $a_n = a + (n - 1) d $$\\$ $= 7 + (8 - 1) 3$ (By Substituting)$ = 7 + (7) 3 = 7 + 21\\ = 28 $Hence, $a_n = 28$

7   Choose the correct choice in the following and justify $\\$ $I.$$30^{th} $ term of the $A.P. 10,7,4,\dots,$ is$\\$ $II. 11^{th} $ term of the $A.P. -3,\frac{-1}{2},2 \dots$

Solution :

I. Given , $\\$ $\bullet $ A.P.Series is $ 10,7,4 ,\dots$$\\$ $\bullet $ First term, $a_1 =10$$\\$ $\bullet $ Second term $ a_2=7$$\\$ $\bullet $ Number of terms ,$ n=30$$\\$ $\bullet $ Common difference ,$ d=a_2-a_1\\ =7-10 \\ =-3$$\\$ $\bullet $ $ 30^{th} $ Term, $a_30 =?$$\\$ We know that the $n^{th}$ term of an A.P. Series,$\\$ $ a_n =a+(n-1)d\\ a_30 =10+(30-1) (-3)$ (By Substituting)$\\$ $a_30 = 10 -87=-77$$\\$ Hence ,the correct answer is $C$$\\$ II.Given that ,$\\$ $\bullet $ A.P. Series is $ -3 ,\frac{-1}{2}, 2\dots $$\\$ $\bullet $ First term $ a=-3 $$\\$ $\bullet $Second Term =$-\frac{1}{2}$$\\$ $\bullet $ Nth term of the A.P. Series , $ a_n =?$$\\$ $\bullet $ Common difference ,$d=a_2 -a_1$$\\$ $ =-\frac{1}{2} -(-3)\\ = -\frac{5}{2}$$\\$ We know that the $n^{th} $ term ofanA.P. Series ,$\\$ $ a_n =a+(n-1)d \\ a_n = -3 +(11-1)(\frac{5}{2}) \\ a_n = -3 +25\\ a_n 22$$\\$ Hence ,the answer is $B$

8   In the following APs find the missing term in the boxes $\\$ $ I. 2,\Box ,26 \\ II. \Box ,13 ,\Box, 3\\ III.5,\Box ,\Box ,9 \frac{1}{2}\\ IV. -4,\Box,\Box,\Box,\Box ,6\\ V.\Box,38,\Box,\Box ,\Box ,-22$

Solution :

On subtracting Equation (I) from Equation (II), we obtain$\\$ $a+d - (a + 3d) = 13-3\\ a + d - a - 3d = 10 \\ -2d =10 d = -5 $$\\$ By Substituting $d = -5$ in equation (I), we obtain$\\$ $ 13 = a + (-5) \\ a = 18\\ a_3 = 18 + (3 - 1) (-5)\\ = 18 + 2 (-5) \\ = 18 - 10 = 8$ $\\$ Therefore, the missing terms are $18$ and $8$ respectively.$\\$ $III. 5,\Box ,\Box , 9\frac{1} {2}$$\\$ For this $A.P.$, $\\$ $\bullet $ First Term , $a= 5$$\\$ $\bullet $Fourth term $ a_4 =9\frac{1}{2}=\frac{19}{2}$$\\$ We know that the nth term of an A.P. Series, $\\$ $ a_n =a+(n+1)d\\ a_4 =a+(4-1)d\\ \frac{19}{2}=5+3d\\ \frac{9}{2}=3d\\ d=\frac{3}{2}\\ a_2=a+d =5+\frac{3}{2}=\frac{13}{2}\\ a_3 = a+2d =5+2(\frac{3}{2}) =8$$\\$ Therefore , the missing term are $\frac{13}{2}$ and $8$ respectively

$IV. -4,\Box ,\Box ,\Box ,\Box , 6$$\\$ For this A.P., $\\$ $\bullet $ First Term,$a = -4$$\\$ $\bullet $ Sixth Term, $a_6 = 6 $$\\$ $\bullet $ Common Difference,$ d=? $$\\$ $\bullet $ Second term $a_2 =? $$\\$ $\bullet $ Third Term, $a_3 = ? $$\\$ $\bullet $ Fourth Term, $a_4 =? $$\\$ $\bullet $ Fifth Term, $a_5 =?$$\\$ We know that the nth term of an A.P. Series,$\\$ $ a_n = a + (n - 1) d\\ a_6 = a + (6 - 1) d \\ 6 = - 4 + 5_d (By Substitutuing) \\ 10 = 5_d \\ d = 2 $$\\$ Hence,$ a_2 = a + d = - 4 + 2 = -2\\ a_3 = a + 2_d = - 4 + 2 (2) = 0\\ a_4 = a + 3_d = - 4 + 3 (2) = 2\\ a_5 = a + 4_d = - 4 + 4 (2) = 4 $$\\$ Therefore, the missing terms are $-2, 0, 2,$ and $ 4 $respectively.$\\$

$II. \Box ,13,\Box , 3$$\\$ For this A.P.,$\\$ Given $\bullet $ Second Term,$a_2=13 $$\\$ $\bullet $ Fourth Term, $a_4 = 3 $$\\$ $\bullet $ First Term,$a =?$$\\$ $\bullet $ Fourth term ,$a_3=?$$\\$ We know that the $n^{th}$ term of an A.P.Series,$\\$ $a_n =a+(n-1)d\\ a_2=a+(2-1)d\\ 13=a+d \dots $$\ \ \ $ Equation (i)$\ \ \ \\$ $a_4 =a+(4-1)d\\ 3=a+3d \dots $$\ \ \ \ $ Equation(ii)$\ \ \ \\$ On subtracting Equation (I) from Equation (II), we obtain$\\$ $a+d - (a + 3d) = 13-3\\ a + d - a - 3d = 10 \\ -2d =10 d = -5 $$\\$ By Substituting $d = -5$ in equation (I), we obtain$\\$ $ 13 = a + (-5) \\ a = 18\\ a_3 = 18 + (3 - 1) (-5)\\ = 18 + 2 (-5) \\ = 18 - 10 = 8$ $\\$ Therefore, the missing terms are $18$ and $8$ respectively.$\\$ $III. 5,\Box ,\Box , 9\frac{1} {2}$$\\$ For this $A.P.$, $\\$ $\bullet $ First Term , $a= 5$$\\$ $\bullet $Fourth term $ a_4 =9\frac{1}{2}=\frac{19}{2}$$\\$ We know that the nth term of an A.P. Series, $\\$ $ a_n =a+(n+1)d\\ a_4 =a+(4-1)d\\ \frac{19}{2}=5+3d\\ \frac{9}{2}=3d\\ d=\frac{3}{2}\\ a_2=a+d =5+\frac{3}{2}=\frac{13}{2}\\ a_3 = a+2d =5+2(\frac{3}{2}) =8$$\\$ Therefore , the missing term are $\frac{13}{2}$ and $8$ respectively

$III. 5,\Box ,\Box , 9\frac{1} {2}$$\\$ For this $A.P.$, $\\$ $\bullet $ First Term , $a= 5$$\\$ $\bullet $Fourth term $ a_4 =9\frac{1}{2}=\frac{19}{2}$$\\$ We know that the nth term of an A.P. Series, $\\$ $ a_n =a+(n+1)d\\ a_4 =a+(4-1)d\\ \frac{19}{2}=5+3d\\ \frac{9}{2}=3d\\ d=\frac{3}{2}\\ a_2=a+d =5+\frac{3}{2}=\frac{13}{2}\\ a_3 = a+2d =5+2(\frac{3}{2}) =8$$\\$ Therefore , the missing term are $\frac{13}{2}$ and $8$ respectively

$I. 2,\Box , 26$ Given, $\bullet a=2 \\ \bullet d=? $$\\$ Let the above term be$ a$, $a+d$ and $a+2d$$\\$ Hence Third Term,$ a+2d = 26$ By Substituting the value of a in the above, $26 = 2 + 2d \\24 = 2d\\ d = 12$ Hence Missing term,$ a+d = 2+12 = 14 $$\\$ $II. \Box ,13,\Box , 3$$\\$ For this A.P.,$\\$ Given $\bullet $ Second Term,$a_2=13 $$\\$ $\bullet $ Fourth Term, $a_4 = 3 $$\\$ $\bullet $ First Term,$a =?$$\\$ $\bullet $ Fourth term ,$a_3=?$$\\$ We know that the $n^{th}$ term of an A.P.Series,$\\$ $a_n =a+(n-1)d\\ a_2=a+(2-1)d\\ 13=a+d \dots $$\ \ \ $ Equation (i)$\ \ \ \\$ $a_4 =a+(4-1)d\\ 3=a+3d \dots $$\ \ \ \ $ Equation(ii)$\ \ \ \\$ On subtracting Equation (I) from Equation (II), we obtain$\\$ $a+d - (a + 3d) = 13-3\\ a + d - a - 3d = 10 \\ -2d =10 d = -5 $$\\$ By Substituting $d = -5$ in equation (I), we obtain$\\$ $ 13 = a + (-5) \\ a = 18\\ a_3 = 18 + (3 - 1) (-5)\\ = 18 + 2 (-5) \\ = 18 - 10 = 8$ $\\$ Therefore, the missing terms are $18$ and $8$ respectively.$\\$ $III. 5,\Box ,\Box , 9\frac{1} {2}$$\\$ For this $A.P.$, $\\$ $\bullet $ First Term , $a= 5$$\\$ $\bullet $Fourth term $ a_4 =9\frac{1}{2}=\frac{19}{2}$$\\$ We know that the nth term of an A.P. Series, $\\$ $ a_n =a+(n+1)d\\ a_4 =a+(4-1)d\\ \frac{19}{2}=5+3d\\ \frac{9}{2}=3d\\ d=\frac{3}{2}\\ a_2=a+d =5+\frac{3}{2}=\frac{13}{2}\\ a_3 = a+2d =5+2(\frac{3}{2}) =8$$\\$ Therefore , the missing term are $\frac{13}{2}$ and $8$ respectively

$V. \Box , 38,\Box ,\Box ,\Box , -22 $$\\$ For this A.P., $ \bullet $ Second term, $a_2 = 38$$\\$ $ \bullet $ Sixth Term, $a_6 = -22 $$\\$ $ \bullet $ First Term, $a=? $$\\$ $ \bullet $ Third Term, $a_3 =?$$\\$ $ \bullet $ Fourth Term, $a_4 = ? $$\\$ $ \bullet $ Fifth Term, $a_5=? $$\\$ $ \bullet $ Common Difference, $d = ?$$\\$ We know that the $n^{th}$ term of an A.P. Series, $\\$ $a_n = a + (n -1) d \\ a_2 = a + (2 - 1) d \\ 38 = a + d \dots $ Equation (1) $\\$ $a_6 = a + (6 - 1) d \\ -22 = a + 5d \dots $ .Equation (2) $\\$ On subtracting equation (1) from Equation (2), we obtain$\\$ $-22 -38 = 4d -60 = 4d\\ d = -15 $$\\$ Hence,$a = a_2 - d = 38 - (-15) = 53\\ a_3 = a + 2d = 53 + 2 (-15) = 23\\ a_4 = a + 3d = 53 + 3 (-15) = 8\\ a_5 = a + 4d = 53 + 4 (-15) = -7$$\\$ Therefore, the missing terms are $53, 23, 8,$ and $-7 $respectively

9   Which term of the $A.P. 3, 8, 13, 18, \dots $is $78$?

Solution :

Given Series, $3, 8, 13, 18, \dots $ For this A.P., $\\$ $\bullet $ First Term,$ a = 3$$\\$ $\bullet $ Common difference, $d = a_2 -a_1 = 8 - 3 = 5 $$\\$ $\bullet $ $n^{th}$ term of this A.P.$a_n = 78 $$\\$ $\bullet $ Sixteenth term, $a_16 = ?$$\\$ We know that for the AP series $n^{th}$ term,$\\$ $a_n = a + (n - 1) d \\ 78 = 3 + (n - 1) 5 (By Substituting) \\ 75 = (n - 1) 5\\ (n - 1) = 15 \\ n = 16$$\\$ Hence, $16^{th}$ term of this A.P. is $78$.

10   Which term of the $A.P. 3, 8, 13, 18, \dots $ is $78$?

Solution :

Given Series, $3, 8, 13, 18, \dots $ For this A.P., $\\$ $ \bullet $ First Term, $a = 3$$\\$ $ \bullet $ Common difference, $d = a_2 - a_1 = 8 - 3 = 5 $$\\$ $ \bullet $$ n^{th}$ term of this A.P. $a_n = 78 $$\\$ $ \bullet $ Sixteenth term, $a_16 = ? $$\\$ We know that for the AP series $n^{th}$ term, $a_n = a + (n - 1) d \\ 78 = 3 + (n - 1) 5 (By Substituting)\\ 75 = (n - 1) 5 (n - 1) = 15\\ n = 16$$\\$ Hence, $16^{th}$ term of this A.P. is $78$.

11   Find the number of terms in each of the following A.P. $\\$ $I. 7, 13, 19, \dots , 205\\ II. 18,15\dfrac{1}{2} ,13.\dots ,-47 $

Solution :

$I. 7, 13, 19, \dots, 205 $$\\$ For this A.P., $\\$ $\bullet $ First term,$a = 7$$\\$ $\bullet $ $n^{th}$ term of the AP series, $a_n = 205$$\\$ $\bullet $ Common Difference, $d = a_2 - a_1 = 13 - 7 = 6 $$\\$ $\bullet $ Number of terms, $n=?$$\\$ We know that the nth term of an A.P. Series,$\\$ $a_n = a + (n - 1) d $$\\$ Therefore, $205 = 7 + (n - 1) 6 \\ 198 = (n - 1) 6\\ 33 = (n - 1)\\ n = 34 $$\\$ Therefore, this given series has $34$ terms in it. $\\$ $II. 18,15\dfrac{1 }{2},13 \dots ,- 47 $$\\$ For this A.P., $ \bullet $ First Term, $a=18$$\\$ $ \bullet $ Common Difference $d = a_2 - a_1 $$\\$ $ 15\dfrac{1}{2} -18 =\dfrac{31}{2}-18\\ \dfrac{31-36}{2}\\ \dfrac{-5}{2}$$\\$ $n^{th}$ term of AP series, $a_n = -47$$\\$ We know that the nth term of an A.P. Series,$\\$ $ a_n =a+(n-1)d\\ -47=18+(n-1)(-\dfrac{5}{2})\\ -65 =(n-1)(-\dfrac{5}{2})\\ (n-1)=\dfrac{-130}{-5} \\ (n-1) =26\\ n=27$$\\$ Therefore ,this given A.P. has $27$ terms in it

12   Check whether $-150$ is a term of the $A.P. 11, 8, 5, 2, \dots $

Solution :

For this $A.P.,$$\\$ $\bullet $ First term, $a = 11$$\\$ $\bullet $ Common difference,$ d = a_2 - a_1 = 8 -11 = -3 $$\\$ $\bullet $ N th term of AP series, $a_n = -150$$\\$ We know that the nth term of an A.P. Series $\\$ $a_n=a+(n-1) d\\ -150 =11+(n-1)(-3)\\ -150=11-3n+3\\ -164=-3n\\ n=\dfrac{164}{3}$$\\$ Clearly, $n$ is not an integer.$\\$ Therefore, $-150$ is not a term of this A.P.

13   Find the $31^{st}$ term of an A.P. whose $11^{th}$ term is $38$ and the $16^th$ term is $73 $

Solution :

Given that, $\\$ $\bullet $ $11^{th}$ term, $a_11 = 38 $$\\$ $\bullet $ $16^{th}$ Term,$a^{16} = 73 $$\\$ $\bullet $ Common difference, $d=?$$\\$ $\bullet $ $31^{st}$ term,$a_31 =?$$\\$ We know that the $n^{th}$ term of an A.P. Series, $\\$ $a_n = a + (n -1) d\\ a_11 = a + (11 - 1) d\\ 38 = a + 10d\dots $ Equation (1) $\\$ Similarly, $a_16 = a + (16 -1) d \\ 73 = a + 15d \dots $ Equation (2) $\\$ On subtracting Equation (1) from Equation (2), we obtain$\\$ $A + 15d - (a +10d) = 73-38 \\ a - a+15d - 10d = 35\\ 5d = 35\\ d = 7 $$\\$ By substituting the value of d in Equation (1), $\\$ $38 = a + 10 × (7) \\ 38 - 70 = a\\ a =-32 \\ a_31 = a + (31 - 1) d \\ = - 32 + 30 (7)\\ = - 32 + 210 \\ = 178$$\\$ Hence,$ 31^{st}$ term is $178$.

14   An A.P. consists of $50$ terms of which $3^{rd}$ term is $12$ and the last term is $106$. Find the $29^{th}$ term

Solution :

15   If $17^{th}$ term of an A.P. exceeds its $10^{th}$ term by $7$. Find the common difference.

Solution :

Given,$\\$ $\bullet a 17 - a \\ 10 = 7$$\\$ $\bullet $ Common difference, $d=?$$\\$ We know that nth term of the AP series,$\\$ For an A.P., $a_ n = a + (n - 1) d\\ a _17 = a + (17 -1) d\\ a _17 = a + 16d$$\\$ Similarly,$ a _10 = a + 9d$$\\$ It is given that $(a + 16d) - (a + 9d) = 7\\ 7d = 7\\ d = 1$$\\$ Therefore, the common difference is $1$.

16   Which term of the $A.P. 3, 15, 27, 39, \dots $ will be $132$ more than its $54^{ th}$ term?

Solution :

17   Two $AP$s have the same common difference. The difference between their $100^{th}$ term is $100$, what is the difference between their $1000^{th}$ terms?

Solution :

Let the first term of these A.P.s be $a_1$ and $b_1$ respectively$\\$ and the Common difference of these A.P’s be $d$. For first A.P.,$\\$ $a_100 = a_1 + (100 - 1) d\\ = a_1 + 99d\\ a_1000 = a_1 + (1000 -1) d\\ a_1000 = a_1 + 999d\\ For second A.P., b_100 = b_1 + (100 -1) d\\ = b_1 + 99d\\ b_1000 = b_1 + (1000 - 1) d\\ =b_1 + 999d$$\\$ Given that, difference between $100^{th}$ term of these $A.P.s = 100$$\\$ Thus, we have $(a_1 + 99d) -(b_1 + 99d) = 100\\ a_1 -b_1 = 100 \ \ \ \ $ Equation (1)$\\$ Difference between $1000^{th}$ terms of these A.P.s$\\$ $(a_1 + 999d) - (b_1 + 999d) \\ = a_1 - b_1 \ \ \ \ $ Equation (2)$\\$ From equation (1) & Equation (2),$\\$ This difference, $a_1 - b_1 = 100$$\\$ Hence, the difference between $1000^{th}$ terms of these A.P. will be $100$.

18   How many three digit numbers are divisible by $7$

Solution :

First three-digit number that is divisible by $7 = 105$$\\$ Next number = $105 + 7 = 112$$\\$ Therefore, the series becomes $105, 112, 119, \dots $$\\$ All are three-digit numbers which are divisible by $7$ and thus,$\\$ all these are terms of an A.P. having $\\$ first term as $105$ and common difference as $7$. When we divide $999$ by $7$, the remainder will be $5$.$\\$ Clearly, $999 - 5 = 994$ is the maximum possible three-digit number that is divisible by $7$. Hence the final series is as follows:$\\$ $105, 112, 119, \dots , 994$$\\$ Let $994$ be the nth term of this A.P.$\\$ $a = 105\\ d = 7\\ a_n = 994\\ n = ?$$\\$ We know that the $n^{th}$ term of an A.P. Series,$\\$ $a_n = a + (n - 1) d\\ 994 = 105 + (n - 1) 7\\ 889 = (n -1) 7\\ (n - 1) = 127\\ n-1 = 889/7\\ n= 127+1\\ n = 128$$\\$ Therefore, $128$ three-digit numbers are divisible by $7$.

19   How many multiples of $4$ lie between $10$ and $250$?

Solution :

By Observation, First multiple of $4$ that is greater than $10$ is $12$.$\\$ Next will be $16$.$\\$ Therefore, The series will be as follows:$12, 16, 20, 24,\dots $$\\$ All these are divisible by $4$ and thus,$\\$ all these are terms of an A.P.$\\$ with first term as $12$ and common difference as $4$.$\\$ When we divide $250$ by $4$,$\\$ the remainder will be $2$. $\\$ Therefore, $250 - 2 = 248$ is divisible by $4$$\\$ which is the largest multiple of $4$ within $250$.$\\$ Hence the final series is as follows:$ 12, 16, 20, 24, \dots, 248$$\\$ Let $248$ be the $n^{th}$ term of this A.P.$\\$ We know that the n th term of an A.P. Series,$\\$ $a = 12\\ d = 4\\ a_n = 248\\ a_n = 1+(n-1)d\\ 248 =12+(n-1)4\\ \dfrac{236}{4}=n-1\\ n=60$$\\$ Therefore, there are$60$ multiples of $4$ between $10$ and $250$.

20   For what value of $n$, are the nth terms of two $APs 63, 65, 67,$ and $3, 10, 17, \dots$ equal

Solution :

If $n^{th}$ terms of the two APs are $63, 65, 67,\dots$$\\$ and $3, 10, 17, \dots $ are equal.$\\$ Then,$ 63 + (n - 1) 2\\ = 3 + (n - 1) 7 .\ \ \ $.Equation (1)$\\$ [Since In $1_st AP, a = 63, d = 65 - 63 = 2$$\\$ and in 2nd AP , $a = 3, d = 10 - 3 = 7]$$\\$ By Simplifying Equation (1)$\\$ $7(n - 1) -2 (n -1) = 63 - 3\\ 7n-7-2n+2 =60\\ 5n -5 = 60\\ n-1 = 12\\ n = 12 + 1 = 13$$\\$ Hence, the $13^{th}$ terms of the two given APs are equal.

21   Determine the A.P. whose third term is $16$ and the $7^{th}$ term exceeds the $5^{th}$ term by $12$.$\\$

Solution :

Let $a$ be the first term and $d$ the common difference.$\\$ Hence from given,$ a_3 = 16$ and $a_7 - a_5 = 12$$\\$ $a + (3 - 1) d = 16\\ a + 2d = 16 \ \ \ \ \dots $Equation (1)$\\$ Using $ a_7 - a_5 = 12\\ [a+ (7 -1) d] - [a + (5 - 1) d] = 12\\ (a + 6d) -(a + 4d) = 12\\ 2d = 12\\ d = 6$$\\$ By Substituting this in Equation (1), we obtain$\\$ $a + 2 (6) = 16\\ a + 12 = 16\\ a = 4$$\\$ Therefore, A.P. will be $4+6,4+2*6,4+3*6 \dots $$\\$ Hence the series will be $4, 10, 16, 22, \dots $

22   Find the $20^{th}$ term from the last term of the $A.P. 3, 8, 13,\dots , 253$

Solution :

Given $A.P. is 3, 8, 13,\dots , 253$$\\$ From Given,$\\$ As the $20^{th}$ term is considered from last $a=253$$\\$ Common difference , $d= 3-8 = -5 \ \ \ $(Considered in reverse order)$\\$ We know that the n th term of an A.P. Series,$\\$ $a_n = a + (n - 1) d$$\\$ Hence $20^{th}$ Term ,$\\$ $ a_20 = a + (20 - 1) d\\ a_20 = 253 + (20 -1) (-5)\\ = 253 - 19 X 5\\ = 253- 95\\ =158$$\\$ Therefore, $20^{th}$ term from the last term is $158$.

23   The sum of $4^{th}$ and $8^{th}$ terms of an A.P. is $24$ and the sum of the $6^{th} and $10^{th}$ terms is $44$. Find the first three terms of the A.P.

Solution :

24   Subba Rao started work in $1995$ at an annual salary of Rs $5000$ and received an increment of Rs $200$ each year. In which year did his income reach Rs $7000$?

Solution :

From the Given Data, $\\$ incomes received by Subba Rao in the years $1995,1996,1997 \dots $$\\$ are $5000, 5200, 5400, \dots 7,000$$\\$ From Observation, Common difference, $d= 200$$\\$ $a = 5000\\ d = 200$$\\$ Let after $n^{th}$ year, his salary be Rs $7000$.$\\$ Hence $a_n = Rs.7000$$\\$ We know that the $n^{th}$ term of an A.P. Series,$\\$ $a_n = a + (n - 1) d$$\\$ By Substituting above values,$\\$ $7000 = 5000 + (n - 1) 200\\ 200(n - 1) = 7000-5000\\ 200(n-1)=2000\\ (n -1) = \dfrac{2000}{200} = 10\\ n = 11$$\\$ Therefore, in $11^{th}$ year, his salary will be Rs $7000$.

25   Ramkali saved Rs $5$ in the first week of a year and then increased her weekly saving by Rs $1.75$. If in the $n^{th}$ week, her weekly savings become Rs $20.75$, find $n.$

Solution :

From the given data, Ramkali’s savings in the consecutive weeks are$\\$ $Rs .5, Rs .5 + Rs .1.75,\\ Rs .5 +2*Rs.1.75,Rs.5+3*Rs.1.75\dots $$\\$ Hence $n^{th}$ Term is $Rs. 5+(n-1) × Rs. 1.75 = Rs. 20.75$$\\$ Now from the above we know that$\\$ $a = 5\\ d = 1.75\\ a_n = 20.75\\ n = ?$$\\$ We know that the n th term of an A.P. Series, $\\$ $a_n=a+(n-1)d\\ 20.75=5+(n-1)1.75\\ 15.75=(n-1)1.75\\ (n-1)=\dfrac{15.75}{1.75}=\dfrac{1575}{175}\\ =\dfrac{63}{7}=9$$\\$ $ n - 1 = 9\\ n = 10$$\\$ Hence, $n$ is $10$.

26   Find the sum of the following APs.$\\$ $(i) 2, 7, 12 ,\dots ,$ to $10$ terms.$\\$ $(ii) - 37, - 33, - 29 ,\dots ,$ to $12$ terms $\\$ $(iii) 0.6, 1.7, 2.8 ,\dots ,$ to $100$ terms $\\$ $(iv) \frac{1}{15},\frac{1}{12},\frac{1}{10},\dots $ to $11$ terms$\\$

Solution :

$(iii) 0.6, 1.7, 2.8 ,\dots ,$ to $100$ terms$\\$ Given,$\\$ $\bullet $ First term,$ a = 0.6$$\\$ $\bullet $ Common difference, $d = a _2 - a _1 = 1.7 - 0.6 = 1.1$$\\$ $\bullet $ Number of Terms,$ n = 100$$\\$ $\bullet $ Sum of $n$ terms, $S _100 =?$$\\$ We know that Sum of $n^{ th}$ term of AP,,$\\$ $S_n=\dfrac{n}{2}[2a+(n-1)d]\\ S_100 =\dfrac{100}{2}[2(0.6)+(100- 1)1.1]\\ =50[1.2+(99)*(1.1)]\\ =6[1.2+108.9]\\ =5505$$\\$ $(iv)\dfrac{1}{15},\dfrac{1}{12},\dfrac{1}{10},\dots $ to $11$ terms for this A.P.,$\\$ Given,$\\$ $\bullet $ First term, $a = \dfrac{1}{15}$$\\$ $\bullet $ Number of Terms,$n=11$$\\$ $\bullet $ Common difference, $d= a _2 -a _1 = \dfrac{1}{12} - \dfrac{1}{15}\\ =\dfrac{ (5-4)}{60}= \dfrac{1}{60}$$\\$ $\bullet $ Sum of $n$ terms, $S _11 =?$ We know that Sum of$n^{th}$ term of series,$\\$ $ S_n=\dfrac{n}{2}[2a +(n-1)d] \\ S_11 =\dfrac{11}{2}[2(\dfrac{1}{15})+(11-1)\dfrac{1}{60}]\\ =\dfrac{11}{2}[\dfrac{2}{15}+\dfrac{10}{60}]\\ =(\dfrac{11}{2})(\dfrac{9}{30})=\dfrac{33}{20}$

$(i)2, 7, 12 ,\dots ,$ to $10$ terms $\\$ Given,$\\$ $\bullet $ First term, $a = 2$$\\$ $\bullet $ Common Difference, $d = a_2 - a_1 =7- 2 = 5$$\\$ $\bullet $ Number of Terms, $n = 10$$\\$ $\bullet $ Sum of $n$ terms, $S_10 =?$$\\$ We know that Sum of $n^{th}$ term of AP,$\\$ $ S_n=\dfrac{n}{2}[2a+(n-1)d]\\ S_10=\dfrac{10}{2}[2(2)+(10-1)5]\\ = 5[4+(9)*(5)]\\ 5*49 = 245$$\\$ $(ii) -37, -33, -29 ,\dots ,$ to $12$ terms$\\$ Given,$\\$ $\bullet $ First term, $a = -37$$\\$ $\bullet $ Common Difference, $d = a_2 - a_1\\ = (-33) - (-37) \\ = -33 + 37 = 4$$\\$ $\bullet $ Number of Terms,$ n = 12$$\\$ $\bullet $ Sum of$n$ terms, $S_12 =?$$\\$ We know that Sum of $n^{th}$ term of AP,$\\$ $S_n=\dfrac{n}{2}[2a+(n-1)d]\\ S_12 =\dfrac{12}{2}[2(-37)+(12-1)4]\\ =6[-74+11*4]\\ =6(-30)\\ =-180$$\\$

27   Find the sums given below$\\$ $(i) 7 + 10\dfrac{1}{2} + 14 +\dots + 84\\ (ii) 34 + 32 + 30 + \dots + 10\\ (iii) -5 + (- 8) + (- 11) + \dots + (- 230)$

Solution :

$(i)7 + 10\dfrac{1}{2} + 14 +\dots + 84$$\\$ Given,$\\$ $\bullet $ First term, $a = 7$$\\$ $\bullet $ Common difference, $d= 10\dfrac{1}{2} - 7 = 7/2$$\\$ $\bullet $ Number of terms, $n =?$$\\$ $\bullet $$ 84^{ th}$ Term, $l= a _n = 84$$\\$ $\bullet $Sum of $n$ terms, $S _23 =?$$\\$l = a n = a + (n - 1)d $l=a_n=a+(n-1)d\\ 84= 7+(n-1)\dfrac{7}{2}\\ 22 = n - 1 Hence, n = 23$ We know that Sum of n terms,$\\$ $S_n =\dfrac{n}{2}[a+1]\\ S_n=\dfrac{23}{2}[7+84]\\ =\dfrac{23*91}{2}=1046\dfrac{1}{2}$ $\\$ $(ii) 34 + 32 + 30 +\dots + 10$$\\$ For this A.P., Given,$\\$ $\bullet $First term,$a = 34$$\\$ $\bullet $ Common Difference, $d = a _2 - a_ 1 = 32 -34 = -2$$\\$ $\bullet $ Last term,$ l = 10 = a _n$$\\$ $\bullet$ Number of terms, $n =?$ $\bullet $ Sum of $n$ terms, $S_10 =?$ We know that the Term of AP,$\\$ $l = a + (n - 1) d\\ 10 = 34 + (n - 1) (-2)\\ -24 = (n - 1) (-2)\\ 12 = n - 1\\ Hence, n = 13$ Sum of n terms of AP is,$\\$ $S_n=\dfrac{n}{2}[a+1]\\ =\dfrac{13}{2}[34+10]\\ =\dfrac{13*44}{2}=13*22=286$$\\$ $(iii) (-5) + (-8) + (-11) +\dots + (-230)$ For this A.P.,$\\$ Given,$\\$ $\bullet $First term, $a = -5$$\\$ $\bullet $ N th Term, $l = -230$$\\$ $\bullet $ Common difference,$ d = a _2 - a _1 = (-8) - (-5)= -8 + 5 = -3$$\\$ $\bullet $ Number f terms, $n =?$$\\$ $\bullet $ Sum of n terms, $S _10 =?$$\\$ We know that $n^{th}$ term of this A.P.$\\$ $l = a + (n - 1)d\\ -230 = -5 + (n -1) (-3)\\ -225 = (n - 1) (-3)\\ (n -1) = 75\\ Hence,n = 76$ We know that sum of $n$ terms of AP,$\\$ $S_n=\dfrac{n}{2}[a+1]\\ =\dfrac{76}{2}[(-5)+(-230)]\\ =38(-235)\\ -8930$

28   In an AP$\\$ (i) Given $ a = 5, d = 3, a_n = 50,$ find $n$ and $S_n$.$\\$ (ii) Given$ a = 7, a _13 = 35,$ find $d$ and $S _13$ .$\\$ (iii) Given $a _12 = 37, d = 3,$ find $a$ and $S _12 .$$\\$ (iv) Given $a _3 = 15, S _10 = 125,$ find $d$ and $a _10 .$$\\$ (v) Given$d = 5, S _9 = 75,$ find $a$ and $a _9 $$\\$. (vi) Given $a = 2, d = 8, S _n = 90$, find $n$ and $a_n$.$\\$ (vii) Given $a = 8, a _n = 62, S_ n = 210,$ find $n$ and $d.$$\\$ (viii) Given $a_ n = 4, d = 2, S_ n = - 14$, find $n$ and $a$.$\\$ (ix) Given $a = 3, n = 8, S = 192$, find $d.$ $\\$ (x)Given $l = 28, S = 144$ and there are total $9$ terms. Find $a.$

Solution :

$S_n=\dfrac{n}{2}[2a+(n-1)d]\\ S_10=\dfrac{9}{2}[2a+(9-1)5]\\ 75=\dfrac{9}{2}(2a+40)\\ 3a=25-60\\ a=\dfrac{-35}{3}$$\\$ We know that nth term of the AP series , $ a_n=a+(n-1)d$$\\$ $A_9=a+(9-1)(5)\\ =\dfrac{-35}{3}+8(5)\\ \dfrac{-35+120}{3}=\dfrac{85}{3}$$\\$ (vi) Given that,$\\$ • First term, $a = 2$$\\$ • Common difference, $d = 8$$\\$ • Sum of n terms, $S_ n =90$$\\$ • nth term of AP, $a _n = ?$$\\$ • number of terms,$n=?$$\\$ As,$S_n=\dfrac{n}{2}[2a+(n-1)d]\\ 90=\dfrac{n}{2}[4+(n-1)8]\\ 90 = n [2 + (n - 1)4]\\ 90 = n [2 + 4n - 4]\\ 90 = n (4n - 2) = 4n^ 2 - 2n\\ 4n ^2 - 2n - 90 = 0\\ 4n ^2 - 20n + 18n -90 = 0\\ 4n (n - 5) + 18 (n - 5) = 0\\ (n -5) (4n + 18) = 0$$\\$ Either $n - 5 = 0 or 4n + 18 = 0\\ n = 5 or =-\dfrac{18}{4}=\dfrac{-9}{2}$$\\$ However, n can neither be negative nor fractional. Therefore, $n = 5\\ a_ n = a + (n - 1)d\\ a _5 = 2 + (5 - 1)8\\ = 2 + (4) (8)\\ = 2 + 32 = 34$$\\$

$S_n=\dfrac{n}{2}[a+a_n] \\ S_n =\dfrac{12}{2}[4+37]\\ S_n=6(41)\\ S_n=246$$\\$ (iv) Given that,$\\$ • Third Term, $a _3 = 15$$\\$ • Number of terms,$ n=10$$\\$ • Sum of n terms, $S_ 10 =125$$\\$ • Common difference, $d = ?$$\\$ • 10th Term,$ _a 10 =?$$\\$ We know that nth term of the AP series, $a_ n = a + (n -1)d,$$\\$ $a_ 3 = a + (3 -1)d\\ 15 = a + 2d\dots ---Equation (i)$$\\$ $S_n=\dfrac{n}{2}[2a+(3-1)d]\\ S_10=\dfrac{10}{2}[2a+(10-1)d]\\ 25=2a+9d$$\\$ On multiplying equation (1) by 2, we obtain$\\$ $30 = 2a + 4d\dots $-Equation (iii) On subtracting equation (iii) from Equation (ii), we obtain $-5 = 5d\\ d = -1$$\\$ From equation (i), $15 = a + 2(-1)\\ 15 = a - 2\\ a = 17\\ a _10 = a + (10 - 1)d\\ a _10 = 17 + (9) (-1)\\ a _10 = 17 - 9 = 8$$\\$ (v) Given that,$\\$ • Common Difference, $d = 5$$\\$ • Number of terms,$ n=9$$\\$ • Sum of n terms,$ S_ 10 =75$$\\$ • First term, $a = ?$$\\$ • 9 th term, $a_ 9 = ?$$\\$ As ,

(vii) Given that,$\\$ • First term, $a = 8$$\\$ • Nth term of$ AP, a n = 62$$\\$ • Sum of n terms, $S_ n =210$$\\$ • Common difference, $d = ?$$\\$ • Number of terms, $n=?$$\\$ $ S_n=\dfrac{n}{2}[a+a_n]\\ 210=\dfrac{n}{2}[8+62]\ \ \ $(By Substituiting )$\\$ $n=6$$\\$ We know that nth term of the AP series, $a _n = a + (n - 1)d\\ 62 = 8 + (6 -1)d\\ 62 - 8 = 5d\\ (By Substituiting) 54 = 5d\\ d=\dfrac{54}{5}$$\\$ (viii) Given that,$\\$ • Common difference, $d = 2$$\\$ • Nth terms of AP, $a_ n = 4$$\\$ • Sum of n terms,$ S _n =-14$$\\$ • First term, $a = ?$$\\$ • Number of terms,$ n=?$$\\$ We know that nth term of AP series, $a_ n = a + (n - 1)d\\ 4 = a + (n -1)2\\ 4 = a + 2n- 2(By Substituting)\\ a + 2n = 6$$\\$ $a = 6 - 2n-\dots -Equation (i) S_n=\dfrac{n}{2}[a+a_n]\\ -41=\dfrac{n}{2}[a+4]\\ -28 = n (a + 4)\\ -28 = n (6 - 2n + 4)$$\\$ {From equation (i)} $-28 = n (- 2n + 10)\\ -28 = - 2n ^2 + 10n\\ 2n ^2 - 10n - 28 = 0\\ n ^2 - 5n -14 = 0\\ n 2 - 7n + 2n - 14 = 0\\ n (n - 7) + 2(n - 7) = 0\\ (n -7) (n + 2) = 0$$\\$ Either $n - 7 = 0 or n + 2 = 0\\ n = 7 or n = -2$$\\$ However, n can neither be negative nor fractional.$\\$ Therefore,$ n = 7$$\\$ From equation (i), we obtain $a = 6 - 2n\\ a = 6 - 2(7)\\ = 6 -14=-8

(i) Given that, • First term, $a = 5$$\\$ • Common difference, $d = 3$$\\$ • Nth term of AP series, $a _n = 50$$\\$ • Number of terms, $n=?$$\\$ • Su of n terms, $S _n =?$$\\$ As $a _n = a + (n - 1)d,\\ \therefore 50 = 5 + (n - 1)3\\ 45 = (n - 1)3\\ 15 = n -1\\ =n=16\\ S_n=\dfrac{n}{2}[a+a_n]\\ S_16=\dfrac{16}{2}[5+50]\\ 440$$\\$ (ii) Given that,$\\$ $\bullet $ First term, $a = 7$$\\$ $\bullet $ Number of terms, $n= 13$$\\$ $\bullet $$13 ^{th}$ term of $AP, a _13 = 35$$\\$ $\bullet $Common difference,$ d=?$$\\$ $\bullet $ Sum of $n$ terms, $S _13 =?$$\\$ We know that nth term of the AP series, $a _n = a + (n -1) d,$$\\$ $\therefore a_ 13 = a + (13 - 1) d$(By Substituting)$\\$ $35 = 7 + 12 d\\ 35 - 7 = 12d\\ 28 = 12d\\ d=\dfrac{7}{3}\\ S_n=\dfrac{n}{2}[a+a_n]\\ S_13=\dfrac{n}{2}[a+a_13]\ \ \ \ (By Substituting)$\\$ =\dfrac{13}{2}[7+35]\\ =273$$\\$ (iii) Given that,$\\$ $\bullet $ First term, $a = ?$$\\$ $\bullet $ Common difference, $d = 3$$\\$ $\bullet $ $12^{ th}$ term of AP series, $a _12 = 37$$\\$ $\bullet $Number of terms, $n=12$$\\$ $\bullet $ First term, $a=? $$\\$ $\bullet $ Sum of n terms, $S _12 =?$$\\$ We know that nth term of the AP series, $a _n = a + (n - 1)d,$$\\$ $a _12 = a + (12 -1)3\\ 37 = a + 33\\ a = 4\\ $$\\$

29   How many terms of the $AP. 9, 17, 25 \dots $ must be taken to give a sum of $636?$

Solution :

Given,$\\$ • First Term, $a= 9$$\\$ • Common Difference, $d = a _2 -a _1 = 17 - 9 = 8$$\\$ • Sum of n terms $S_ n =636$$\\$ • Number of terms,$ n= ?$$\\$ We know that sum of n terms of AP $S_n=\dfrac{n}{2}[2a+(n-1)d]\\ 636=\dfrac{n}{2}[18+(n-1)8]\\ 636 = n [9 + 4n - 4]\\ 636 = n (4n + 5)\\ 4n ^2 + 5n -636 = 0\\ 4n ^2 + 53n - 48n - 636 = 0\\ n (4n + 53) - 12 (4n + 53) = 0\\ (4n + 53) (n - 12) = 0\\ Either 4n + 53 = 0 or n -12 = 0\\ n=\dfrac{-53} {4} Or n=12$ $\\$ $n$ cannot be $\dfrac{-53}{4}$ .As the number of terms can neither be negative nor fractional, therefore,$ n = 12 $only.

30   The first term of an AP is $5$, the last term is $45$ and the sum is $400.$ Find the number of terms and the common difference.

Solution :

Given that,$\\$ • First term, $a = 5$$\\$ • Last term,$ l= a _n = 45$$\\$ • Sum of n terms, $S_ n = 400$$\\$ • Number of terms,$ n=?$$\\$ • Common difference,$ d=?$$\\$ We know that sum of n terms,$\\$ $S_n=\dfrac{n}{n}(a+1)\\ 400=\dfrac{n}{2}(5+45)$ (By Substituting)$\\$ $400=\dfrac{n}{2}(50)\\ n = 16\\ l = a + (n - 1) d\\ 45 = 5 + (16 -1) d\\ 40 = 15d\\ d=\dfrac{40}{15}=\dfrac{8}{3}$

31   The first and the last term of an AP are $17$ and $350$ respectively. If the common difference is $9$, how many terms are there and what is their sum?

Solution :

Given that,$\\$ • First term, $a = 17$$\\$ • Last term,$ l = 350$$\\$ • Common difference, $d = 9$$\\$ • Number of terms, $n=?$$\\$ • Sum of n terms, $S_ n =?$$\\$ We know that nth term of $AP, l = a + (n -1) d$$\\$ $350 = 17 + (n - 1)9\\ 333 = (n - 1)9(By Substituting)\\ (n - 1) = 37\\ n = 38$$\\$ Sum of n terms of AP,$\\$ $S_n=\dfrac{n}{2}(a+1)\\ =S_n=\dfrac{38}{2}(17+350)\\ =6973$$\\$ Thus, this A.P. contains 38 terms and the sum of the terms of this A.P. is $6973$.

32   Find the sum of first $22$ terms of an AP in which $d = 7$ and $22^{nd}$ term is $149.$

Solution :

Given,$\\$ • Common Difference,$ d = 7$$\\$ • 22 nd term, $a_ 22 = 149$$\\$ • Sum of n terms, $S_ 22 = ?$$\\$ • $a=?$$\\$ We know that nth term of$AP, a _n = a + (n - 1)d$$\\$ $a _22 = a + (22 - 1)d\\ 149 = a + 21 × 7 (By Substituting)\\ 149 = a + 147\\ a = 2\\ S_n=\dfrac{n}{2}(a+a_n)\\ \dfrac{22}{2}(2+149)\\ 11(151)=1661$

33   Find the sum of first $51$ terms of an AP whose second and third terms are $14$ and $18$ respectively.

Solution :

Given that,$\\$ • First term,$a=?$$\\$ • Second term,$ a_ 2 = 14$$\\$ • Third Term, $a_ 3 = 18$$\\$ • Common difference,$ d = a _3 - a_2 = 18 - 14 = 4$$\\$ • Sum of n terms, =? We know that second term, $a_ 2 = a + d\\ 14 = a + 4\\ a = 10$ Sum of nth term of series,$\\$ $S_n=\dfrac{n}{2}[2a+(n-1)d]\\ S_51=\dfrac{51}{2}[2*10+(51-1)4]\\ \dfrac{51(220)}{2}=51(110)\\ =5610$

34   If the sum of first $7$ terms of an AP is $49$ and that of $17$ terms is $289$, find the sum of first $n$ terms.

Solution :

Given that,$\\$ • Sum on $7$ terms, $S_ 7 = 49$$\\$ • Sum of $17$ terms, $S _17 = 289$$\\$ • Sum of $n$ terms, $S_ n = ?$$\\$ • Number of terms,$ n =?$$\\$ • Common difference, $d=?$$\\$ • First term,$ a=?$$\\$ We know that sum of n term of AP is ,$\\$ $S_n=\dfrac{n}{2}[2a]+(n-1)d]\\ 49=\dfrac{7}{2}[2a+6d]\\ 7=(a+3d)\\ a+3d=7 \ \\ \ \ \(i)$$\\$ Similarly ,$ S_17 =\dfrac{17}{2}[2a+(17-1)d]\\ 289=\dfrac{17}{2}[2a+16d]\\17=(a+8d)\\ a+8d=17 \ \ \ \ \ \ ( ii)$$\\$ Subtracting equation (i) from equation (ii), $5d = 10\\ d = 2$$\\$ From equation (i), $a + 3(2) = 7\\ a + 6 = 7\\ a = 1\\ S_n=\dfrac{n}{2}[2a+(n-1)d]\\ =\dfrac{n}{2}[2(1)+(n-1)(2)]\\ \dfrac{n}{2}(2+2n-2)\\ \dfrac{n}{2}(2n)\\ n^2$

35   Show that $a_1, a_2 \dots , a_n , \dots $ form an AP where an is defined as below $(i) a _n = 3 + 4n\\ (ii) a _n = 9 - 5n$$\\$ Also find the sum of the first $15$ terms in each case.

Solution :

(ii)Given,$\\$ • Nth term of $AP, a _n = 9 - 5n$$\\$ • Common difference,$ d=?$$\\$ • First term, $a=?$$\\$ $a_ 1 = 9 - 5 × 1 = 9 - 5 = 4\\ a _2 = 9 - 5 × 2 = 9 - 10 = -1\\ a _3 = 9 - 5 × 3 = 9 - 15 = -6\\ a _4 = 9 - 5 × 4 = 9 - 20 = -11$$\\$ It can be observed that$\\$ $a _2 - a _1 = - 1 -4 = -5\\ a _3 - a _2 = - 6 - (-1) = -5\\ a _4 - a 3 = - 11 - (-6) = -5$$\\$ i.e., $a_ k + 1 - a_ k$ is same every time.$\\$ Therefore, this is an A.P. with common difference as$ -5$ and first term as $4$. Sum of n terms of AP,$\\$ $ S_n=\dfrac{n}{2}[2a(n-1)d]\\ S_15=\dfrac{15}{2}[2(4)+(15-1)(-5)]\\ \dfrac{15}{2}(8-70)\\ \dfrac{15}{2}(-62)\\= 15(-13)\\ =-465$

(i)Given,$\\$ • Nth term of $AP, a_ n = 3 + 4n$$\\$ • Common difference, $d=?$$\\$ • First term, $a=?$$\\$ • Sum of $15$ terms, $S_ n =?$$\\$ $a_ 1 = 3 + 4(1) = 7\\ a _2 = 3 + 4(2) = 3 + 8 = 11\\ a _3 = 3 + 4(3) = 3 + 12 = 15\\ a _4 = 3 + 4(4) = 3 + 16 = 19$$\\$ It can be observed that$\\$ $a_ 2 -a 1 = 11 - 7 = 4\\ a _3 - a 2 = 15 -11 = 4\\ a_ 4 - a 3 = 19 -15 = 4$$\\$ i.e., $a_ n+ 1$ and $a_ n$ is same every time.$\\$ Therefore, this is an AP with common difference as $4$ and first term as $7$.$\\$ Sum of n terms,$\\$ $S_n=\dfrac{n}{2}[2a+(n-1)d]\\ S_15=\dfrac{15}{2}[2(7)+(15-1)4]\\ \dfrac{15}{2}((14)+6)\\ \dfrac{15}{2}(70)\\ 15*35\\ 525$

36   If the sum of the first n terms of an AP is $4n - n_ 2$ , what is the first term (that is $S_ 1$ )? What is the sum of first two terms? What is the second term? Similarly find the $3_rd$, the $10^{th}$ and the $n^{th}$ terms.

Solution :

Given,$\\$ • Sum of first n terms,$ S _n = 4n - n_ 2$$\\$ • Sum of First term, $a = S _1 = ?$$\\$ • Sum of first two terms =$S _2 =?$$\\$ • Second term,$ a_ 2 = ?$$\\$ • Common Difference, $d = a _2 - a = 1 - 3 = -2$$\\$ • Third term, $a _3 = ?$$\\$ • 10 th term,$ a _10 = ?$$\\$ • Nth term, $a_ n = ?$$\\$ Sum of First term, $a = S _1 = 4(1) -(1)^2 = 4 - 1 = 3$$\\$ Sum of first two terms =$ S_ 2 =?= 4(2) - (2) ^2 = 8 - 4 = 4$$\\$ Second term, $a_ 2 = S _2 - S_ 1 = 4 - 3 = 1$$\\$ We know that sum of n terms, $a_ n = a + (n - 1)d\\ = 3 + (n - 1) (-2)\\ = 3 -2n + 2\\ = 5 - 2n$ Therefore, $a_ 3 = 5 -2(3) = 5- 6 = -1\\ a _10 = 5 - 2(10) = 5 - 20 = -15$$\\$ Hence, the sum of first two terms is $4.$ The second term is $1.$ $3_rd, 10^{th},$ and $n^{th}$ terms are $-1, -15,$ and $5 - 2n$ respectively.

37   Find the sum of first $40$ positive integers divisible by $6$

Solution :

The positive integers that are divisible by $6$ are $6, 12, 18, 24 \dots $$\\$ It can be observed that these are making an A.P. Hence,$\\$ • first term, $a= 6$$\\$ • common difference $,d =6.$$\\$ • Sum of $40$ terms, $S _40 =?$$\\$ As we know that Sum of $n$ terms,$\\$ $ S_n=\dfrac{n}{2}[2a(n-1)d]\\ S_40=\dfrac{40}{2}[2(6)+(40-1)6]\\ = 20[12 + (39) (6)]\\ = 20(12 + 234)\\ = 20 × 246\\ = 4920$

38   Find the sum of first $15$ multiples of$8$.

Solution :

The multiples of $8$ are $8, 16, 24, 32...$ These are in an A.P., Hence,$\\$ • First term, $a = 8$$\\$ • Common difference, $d = 8$$\\$ • Sum of $15$ terms, $S_ 15 =?$$\\$ Therefore,

39   Find the sum of the odd numbers between $0$ and $50.$

Solution :

The odd numbers between $0$ and $50$ are$\\$ $1, 3, 5, 7, 9 \dots 49$$\\$ Therefore, it can be observed that these odd numbers are in an A.P. • First term, $a = 1$$\\$ • Common difference, $d = 2$$\\$ • Last term,$ l = 49$$\\$ • Sum of odd numbers between $ 0$ and $50, S _n =?$$\\$ We know that nth term of $AP, l = a + (n - 1) d$$\\$ $49 = 1 + (n - 1)2\\ 48 = 2(n - 1)\\ n - 1 = 24\\ n = 25$ We know that sum of $n$ terms of AP,$\\$ $ S_n =\dfrac{n}{2}(a+1)\\ S_25 =\dfrac{25}{2}(1+49)\\ =dfrac{25(50)}{2}=(25)(25)\\ =625$

40   A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: $Rs. 200$ for the first day, $Rs. 250$ for the second day,$ Rs. 300$ for the third day, etc., the penalty for each succeeding day being $Rs. 50$ more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by $30$ days.

Solution :

By observation that these penalties are in an A.P. having first term$\\$ as $200$ and common difference as $50.$$\\$ $a = 200\\ d = 50$$\\$ Penalty that has to be paid if he has delayed the work by $30 $days = $S_ 30 =\dfrac{30}{2}[ 2 ( 200 )+(30 - 1 ) 50 ]\\ =15[400+1450]\\ 15(1850)\\ 27750$$\\$ Therefore, the contractor has to pay $Rs 27750$ as penalty.

41   A sum of $Rs 700$ is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is $Rs 20$ less than its preceding prize, find the value of each of the prizes.

Solution :

• Let the cost of $1_{st}$ prize be $P.$$\\$ • Cost of $2^{nd}$ prize = $P - 20$$\\$ • And cost of $3^{rd}$ prize = $P - 40$$\\$ By observation that the cost of these prizes are in an A.P. having common difference as $-20$ and first term as $P.$$\\$ $a = P\\ d = -20$$\\$ Given that, $S _7 = 700\\ \dfrac{7}{2}[2a+(7-1)d]=700\\ \dfrac{[2a+(6)(-20)]}{2}=100\\ a+3(-20)=100\\ a-60=100\\ a=160$$\\$ Therefore, the value of each of the prizes was $Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60,$ and $Rs 40.$

42   In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of class $I$ will plant $1$ tree, a section of class $II$ will plant $2$ trees and so on till class $XII.$ There are three sections of each class. How many trees will be planted by the students?

Solution :

It can be observed that the number of trees planted by the students is in an AP.$\\$ $1, 2, 3, 4, 5.\dots 12$$\\$ • First term,$ a = 1$$\\$ • Common difference, $d = 2 - 1 = 1$$\\$ We know that sum of $n$ terms of AP,$\\$ $S_n=\dfrac{n}{2}[2a+(n-1)d]\\ S_12=\dfrac{12}{2}(2(1)+(12-1)(1))\\ =6(2+11)\\ 6(13)\\ 78$$\\$ Therefore, number of trees planted by $1$ section of the classes = $78$$\\$ Number of trees planted by $3$ sections of the classes = $3 × 78 = 234$$\\$ Therefore, $234$ trees will be planted by the students.

43   A spiral is made up of successive semicircles, with centres alternately at $A$ and $B$, starting with centre at $A$ of radii $0.5, 1.0 cm, 1.5 cm, 2.0 cm, \dots $ as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles?[take $\pi=\dfrac{22}{7}$]

Solution :

Semi-perimeter of circle = $ \pi r\\ I _1 = \pi (0.5) = \dfrac{\pi}{2}cm\\ I _2 = \pi(1) = \pi cm\\ I _3 = \pi (1.5) = \dfrac{3\pi}{2}cm$$\\$ Therefore, $I _1 , I _2 , I _3 $,i.e. the lengths of the semi-circles are in an A.P.,$\\$ $\dfrac{\pi}{2},\pi,\dfrac{3\pi}{2},2\pi, \dots \\ a=\dfrac{\pi}{2}\\ d=\pi-\dfrac{\pi}{2}=\dfrac{\pi}{2}\\ S_13=?$$\\$ We know that the sum of n terms of an a A.P. is given by$\\$ $S_n=\dfrac{n}{2}[2a+(n-1)d]\\ =\dfrac{13}{2}[2(\dfrac{\pi}{2})+(13-1)(\dfrac{\pi}{2})]\\ =\dfrac{13}{2}[\pi+\dfrac{12\pi}{2}]\\ (\dfrac{13}{2})(7\pi)\\ \dfrac{91\pi}{2}\\ \dfrac{91*22}{2*7}=13*11\\ =143$$\\$ Therefore, the length of such spiral of thirteen consecutive semi-circles will be $143 cm.$

44   $200$ logs are stacked in the following manner: $20$ logs in the bottom row, $19$ in the next row, $18$ in the row next to it and so on. In how many rows are the $200$ logs placed and how many logs are in the top row?

Solution :

It can be observed that the numbers of logs in rows are in an A.P. $20, 19, 18\dots $$\\$ For this A.P.,$\\$ • First term, $a = 20 $$\\$ • Common Difference, $d = a_2 - a _1 = 19 - 20 = -1 $$\\$ Let a total of $200$ logs be placed in n rows. • $S_ n = 200 $$\\$ We know that sum of $n$ terms of AP,$\\$ $S_n=\dfrac{n}{2}[2a+(n-1)d]\\ 200=\dfrac{n}{2}(2(20)+(n-1)(-1))\\ 400 = n (40 - n + 1)\\ 400 = n (41 - n)\\ 400 = 41n - n2\\ n_ 2 - 41n + 400 = 0\\ n _2 -16n - 25n + 400 = 0\\ n (n -16) -25 (n - 16) = 0\\ (n - 16) (n - 25) = 0$$\\$ Either $(n -16) = 0 or n - 25 = 0\\ n = 16 or n = 25$$\\$ $a_ n = a + (n -1)d\\ a _16 = 20 + (16 - 1) (-1)\\ a _16 = 20 - 15\\ a _16 = 5$$\\$ Similarly,$ a _25 = 20 + (25 -1) (-1)\\ a_ 25 = 20 - 24\\ = -4$$\\$ Clearly, the number of logs in $16^{th}$ row is $5$. $\\$ However, the number of logs in $25^{th}$ row is negative, which is not possible.$\\$ Therefore, $200$ logs can be placed in $16$ rows and the number of logs in the $16^{th}$ row is $5.$

45   In a potato race, a bucket is placed at the starting point, which is $5 m$ from the first potato and other potatoes are placed $3 m$ apart in a straight line. There are ten potatoes in the line. A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run? [Hint: to pick up the first potato and the second potato, the total distance (in metres) run by a competitor is $2 × 5 + 2 ×(5 + 3)$]

Solution :

The distances of potatoes are as follows.$\\$ $5, 8, 11, 14\dots $ It can be observed that these distances are in A.P. • First term, $a = 5$$\\$ • Common difference, $d = 8 - 5 = 3$$\\$ We know that the sum of $n$ terms,$\\$ $S_n=\dfrac{n}{2}(2a+(n-1)d)\\ S_10=\dfrac{10}{2}(2(5)+(10-1)3)\\ =5[10+9*3]\\ 5(10+27)=185$$\\$ As every time she has to run back to the bucket, therefore, the total distance that the competitor has to run will be two times of it. Therefore, total distance that the competitor will run = $2 × 185 = 370 m$$\\$ $Alternatively,$ The distances of potatoes from the bucket are $ 5, 8, 11, 14\dots $$\\$ Distance run by the competitor for collecting these potatoes are two times of the distance at which the potatoes have been kept. Therefore, distances to be run are $10, 16, 22, 28, 34,\dots \\ a = 10\\ d = 16 - 10 = 6\\ S _10 =?\\ S_10=\dfrac{10}{2}(2*10+(10-1)6)\\ =5[20+54]\\ =370$$\\$ Therefore, the competitor will run a total distance of $370 m.$

46   Which term of the $A.P. 121, 117, 113 \dots $ is its first negative term? [Hint: Find n for $a_ n < 0$]

Solution :

Given:$\\$ • A.P. Series:$ 121, 117, 113 \dots $$\\$ • First Term, $a = 121 $$\\$ • Common Difference,$d = 117 - 121 = -4 $$\\$ • The first negative term of this A.P. Series =?$\\$ We know that n th term of $AP, a _n = a + (n - 1) d $$\\$ By substituting the above values$\\$ $a _n = 121 + (n -1) (-4)\\ = 121 - 4n + 4\\ = 125 - 4n$$\\$ For the first negative term of this A.P, Therefore, $a_n=<0\\ 125-4n<0\\ =125<4n\\ n>\dfrac{125}{4}\\ n>31.25$$\\$ Therefore, $32^{ nd}$ term will be the first negative term of this A.P.

47   The sum of the third and the seventh terms of an A.P is $6$ and their product is $8$. Find the sum of first sixteen terms of the A.P.

Solution :

Given:$\\$ • $a _3 + a_7 = 6 \ \ \ \ $Equation (1)$\\$ • $(a_ 3 ) × (a_ 7 ) = 8 \ \ \ \ \ $Equation (2)$\\$ We know that $n^{th}$ term of AP Series,$\\$ $a_ n = a + (n - 1) d$$\\$ Third Term, $a_ 3 = a + (3 - 1) d$$\\$ Hence, $a_ 3 = a + 2d\ \ \ \ \ $ Equation (3)$\\$ Seventh Term, $a_ 7 = a + 6d\ \ \ \ $Equation (4)$\\$ Using Equation (3) and Equation (4) in Equation (1) ,$\\$ $(a + 2d) + (a + 6d) = 6\\ 2a + 8d = 6\\ a + 4d = 3\\ a = 3 - 4d \ \ \ \ \ $-Equation(5)$\\$ Using Equation (3) and Equation (4) in Equation(2) ,$\\$ $(a + 2d) × (a + 6d) = 8$$\\$ Substituting the value of Equation(5) in above,$\\$ $(3-4d+2d)*(3-4d+6d)=8\\ (3-2d)*(3+2d)=8\\ 9-4d^2=8\\ 4d^2=1\\ d^2=\dfrac{1}{4}\\ d=\pm \dfrac{1}{2}\\ d=\dfrac{1}{2} Or -\dfrac{1}{2}$$\\$ Hence by substituting both the values of d,$\\$ (When d is $\frac{1}{2}$)$\\$ $a=3-4d\\a=3-4(\dfrac{1}{2})\\ =3-2=1$$\\$ (When d is $ -\frac{1}{2}$)$\\$ $a=3-4d\\a=3-4(-\dfrac{1}{2})\\ =3+2=1$$\\$ We know that Sum of Nth term of AP Series,$\\$ $S_n=\dfrac{n}{2}[2a+(n-1)d]$$\\$ (When a is $1$ and d is$\dfrac {1}{2}$)$\\$ $ S_16 =\dfrac{16}{2}[2(1)+(16-1)(\dfrac{1}{2})]\\ =8[2+\dfrac{15}{2}]\\ =76$$\\$ (When a is $5$ and d is $ -\dfrac{1}{2})$$\\$ $8[10+(15)(-\dfrac{1}{2})]\\ 8(\dfrac{5}{2})\\ =20$

48   A ladder has rungs $25 cm $ apart. (See figure). The rungs decrease uniformly in length from $45 cm$ at the bottom to $25 cm$ at the top. If the top and bottom rungs are $ 2\dfrac{ 1}{2 }m$ apart, what is the length of the wood required for the rungs?[Hint : number of rungs =$\dfrac{250}{25}$]

Solution :

Given:$\\$ • Distance between the rungs= $25 cm $$\\$ • Distance between the top and bottom rungs =$2\dfrac{1}{2}m =2\dfrac{1}{2}*100 cm$$\\$ $ \therefore $ Total number of rungs =$\dfrac{2\dfrac{1}{2}*100}{25}+1\\ =\dfrac{250}{25}+1=11$$\\$ From the given Figure, we can observe that the lengths of the rungs decrease uniformly, hence we can conclude that they will be in an AP The length of the wood required for the rungs equals the sum of all the terms of this A.P.$\\$ • First term, $a = 45$$\\$ • Last term,$ l = 25$$\\$ • No of terms, $n = 11$$\\$ Hence Sum of $n^{th}$ Series,$\\$ $ S_n=\dfrac{n}{2}(a+1)\\ \therefore S_10=\dfrac{11}{2}(45+25)\\ 385 cm$$\\$ Therefore, the length of the wood required for the rungs is $385 cm.$

49   The houses of a row are number consecutively from $1$ to $49$. Show that there is a value of $x$ such that the sum of numbers of the houses preceding the house numbered $x$ is equal to the sum of the number of houses following it. Find this value of $x$. [Hint $S_x - 1 = S_49 - S_x$]

Solution :

Given :$\\$ Number of houses was $1,2,3,\dots 49$$\\$ By Observation, the number of houses are in an A.P. Hence$\\$ • First Term, $a= 1$$\\$ • Common Difference, $d =1$$\\$ Let us assume that the number of $x^{ th}$ house can be expressed as below:$\\$ We know that, Sum of $n$ terms in an A.P.$\\$ $\dfrac{n}{2}[2a+(n-1)d]$$\\$ Sum of number of houses preceding $x ^{th}$ house = $S _x -1$$\\$ $=\dfrac{(x-1)}{2}[2a+(x-1-1)d]\\ =\dfrac{x-1}{2}[2(1)+(x-2)(1)]\\ =\dfrac{(x)(x-1)}{2}$$\\$ By the given we know that, Sum of number of houses following xth house = $S_ 49 - S _x$$\\$ $\dfrac{49}{2}(2(1)+(49-1)(1))\\ -\dfrac{x}{2}[2(1)+(x-1)(1)]\\ =(\dfrac{49}{2})(50)-\dfrac{x}{2}(x+1)\\ =25(49)- \dfrac{x(x+1)}{2}$$\\$ It is given that these sums are equal to each other.$\\$ $\dfrac{x(x-1)}{2}=25(49)-x\dfrac{x+1}{2}\\ \dfrac{x^2}{2}-\dfrac{x}{2}=1225 -\dfrac{x^2}{2}-\dfrac{x}{2}\\ x^2=1225\\ x=\pm 35$$\\$ As the number of houses cannot be a negative number we consider number of houses, $x = 35.$$\\$ Therefore, house number $35$ is such that the sum of the numbers of houses preceding the house numbered $35$ is equal to the sum of the numbers of the houses following it.

50   A small terrace at a football ground comprises of $15$ steps each of which is $50 m$ long and built of solid concrete. Each step has a rise of $\dfrac{1}{4} m $ and a tread of $\dfrac{1}{ 2}m$ (See figure) calculate the total volume of concrete required to build the terrace.

Solution :

Given: $\\$ o From the figure, it can be observed that $1^{ st}$ step is $\dfrac{1}{2} m$ wide,$2^{ nd}$ step is $1 m$ wide,$3 ^{rd }$ step is $\dfrac{3}{2} m$ wide. Therefore, the width of each step is increasing by $\dfrac{1}{2} m$ each time height $\dfrac{1}{4}m$$\\$ o length $50 m$ remains the same. Widths of these steps are$ \dfrac{1}{2},1,\dfrac{3}{2},2,\dots $$\\$

Volume of Step can be considered as Volume of Cuboid= Length $X$ Breadth $X$ Height$\\$ Volume of concrete in 1st step=$\dfrac{1}{4}*\dfrac{1}{2}*50=\dfrac{25}{4}$$\\$ Volume of concrete in 2nd step=$\dfrac{1}{4}*1*50=\dfrac{25}{2}$$\\$ Volume of concrete in 3rd step=$\dfrac{1}{4}*\dfrac{3}{2}*50=\dfrac{75}{4}$$\\$ It can be observed that the volumes of concrete in these steps are in An A.P.$\\$ $ \dfrac{25}{4},\dfrac{25}{2},\dfrac{75}{4},\dots $$\\$ First term, $ a=\dfrac{25}{4}\\ d=\dfrac{25}{2}-\dfrac{25}{4}=\dfrac{25}{4}$$\\$ Sum of $n^{th}$ term, and $S_n=\dfrac{n}{2}[2a+(n-1)d]\\ S_15=\dfrac{15}{2}(2(\dfrac{25}{4})+(15-1)\dfrac{25}{4})\\ =\dfrac{15}{2}[\dfrac{25}{2}+\dfrac{175}{2}]\\ =\dfrac{15}{2}(100)=750$$\\$ Volume of concrete required to build the terrace is $750 m^ 3 .$