# 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 :

(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.

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

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 :

(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.

(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.

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 :

$\frac{1}{3},\frac{5}{3},\frac{9}{3},\frac{13}{3}$ $\\$ 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

(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

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 :

(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$$\\ (iv) The given Series is -10, - 6, - 2, 2 \dots$$\\$ 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$$\\ Seventh Term, a_7 = 6a+a =7a$$\\$

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 :

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, $a_n = 28$

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 $$a_n = ? \\ We know that the n^{th } term of an A.P. Series,\\ a_n = a + (n - 1) d \\ a_n = 3.5 + (105 - 1) 0 (By Substituting)\\ a_n = 3.5 + 104 × 0\\ a_n = 3.5 7 Choose the correct choice in the following and justify \\ I.$$30^{th}$ term of the $A.P. 10,7,4,\dots,$ is$\\$ $II. 11^{th}$ term of the $A.P. -3,\frac{-1}{2},2 \dots$

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

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 $$\\ Therefore, the missing terms are -2, 0, 2, and 4 respectively.\\ 9 Which term of the A.P. 3, 8, 13, 18, \dots is 78? ##### Solution : Given Series, 3, 8, 13, 18, \dots For this A.P., \\ \bullet First Term, a = 3$$\\$ $\bullet$ 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. $\\$ $I. 7, 13, 19, \dots , 205\\ II. 18,15\dfrac{1}{2} ,13.\dots ,-47$

$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.

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$$\\ Therefore, the common difference is 1. 16 Which term of the A.P. 3, 15, 27, 39, \dots will be 132 more than its 54^{ th} term? ##### Solution : 17 Two APs have the same common difference. The difference between their 100^{th} term is 100, what is the difference between their 1000^{th} terms? ##### Solution : Let the first term of these A.P.s be a_1 and b_1 respectively\\ and the Common difference of these A.P’s be d. For first A.P.,\\ a_100 = a_1 + (100 - 1) d\\ = a_1 + 99d\\ a_1000 = a_1 + (1000 -1) d\\ a_1000 = a_1 + 999d\\ For second A.P., b_100 = b_1 + (100 -1) d\\ = b_1 + 99d\\ b_1000 = b_1 + (1000 - 1) d\\ =b_1 + 999d$$\\$ 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 .$$\\ (v) Givend = 5, S _9 = 75, find a and a _9$$\\$. (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.$$\\ (viii) Given a_ n = 4, d = 2, S_ n = - 14, find n and a.\\ (ix) Given a = 3, n = 8, S = 192, find d. \\ (x)Given l = 28, S = 144 and there are total 9 terms. Find a. ##### Solution : S_n=\dfrac{n}{2}[a+a_n] \\ S_n =\dfrac{12}{2}[4+37]\\ S_n=6(41)\\ S_n=246$$\\$(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,$$\\ a _12 = a + (12 -1)3\\ 37 = a + 33\\ a = 4\\$$\\$

$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$$\\ 29 How many terms of the AP. 9, 17, 25 \dots must be taken to give a sum of 636? ##### Solution : Given,\\ • First Term, a= 9$$\\$ • 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 $149.$

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$8$.

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 $\pi=\dfrac{22}{7}$]

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 =$\dfrac{250}{25}$]