# 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 $A.P. -3,\frac{-1}{2},2 \dots$

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. $\\$ $I. 7, 13, 19, \dots , 205\\ II. 18,15\dfrac{1}{2} ,13.\dots ,-47$

$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 $158$.