Class 11: Arithmetic Progressions – Exercise 19.4 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Find the sum of the following arithmetic progressions:

i) $50,46, 42, \ldots$ to $10$ terms

ii) $1,3, 5,7, \ldots$ to $12$ terms

iii) $3, 9/2, 6, 15/2, \ldots$ to $25$ terms

iv) $41, 36, 31, \ldots$ to $12$ terms

v) $a + b, a -b, a - 3b, \ldots$ to $22$ terms

vi) $(x-y)^2, (x^2+y^2) , (x+y)^2, \ldots$ to $n$ terms

vii) $\frac{x-y}{x+y}, \frac{3x-2y}{x+y}, \frac{5x-3y}{x+y}, \ldots$ to $n$ terms

Answer:

i) Given series $50,46, 42, \ldots$

Given $a = 50, \ \ \ \ \ d = (46-50) = -4 \ \ \ \ \ n=10$

We know $S_n =$ $\frac{n}{2}$ $\Big[ 2a + ( n - 1) d \Big]$

$\therefore S_{10} =$ $\frac{10}{2}$ $\Big[ 2(50) + ( 10 - 1) (-4) \Big] = 5 (100-36) = 320$

ii) Given series $1,3, 5,7, \ldots$

Given $a = 1, \ \ \ \ \ d = (3-1) = 2 \ \ \ \ \ n=12$

We know $S_n =$ $\frac{n}{2}$ $\Big[ 2a + ( n - 1) d \Big]$

$\therefore S_{12} =$ $\frac{12}{2}$ $\Big[ 2(1) + ( 12 - 1) (2) \Big] = 6 (24) = 144$

iii) Given series $3, 9/2, 6, 15/2, \ldots$

Given $a = 3, \ \ \ \ \ d = ($ $\frac{9}{2}$ $-3) =$ $\frac{3}{2}$ $\ \ \ \ \ n=25$

We know $S_n =$ $\frac{n}{2}$ $\Big[ 2a + ( n - 1) d \Big]$

$\therefore S_{25} =$ $\frac{25}{2}$ $\Big[ 2(3) + ( 25 - 1) ($ $\frac{3}{2}$ $) \Big] =$ $\frac{25}{2}$ $\times 42 = 525$

iv) Given series $41, 36, 31, \ldots$

Given $a = 41, \ \ \ \ \ d = (36-41) = -5 \ \ \ \ \ n=12$

We know $S_n =$ $\frac{n}{2}$ $\Big[ 2a + ( n - 1) d \Big]$

$\therefore S_{12} =$ $\frac{12}{2}$ $\Big[ 2(41) + ( 12- 1) (-5) \Big] = 6 \times 27 = 162$

v) Given series $a + b, a -b, a - 3b, \ldots$

Given $a = a+b, \ \ \ \ \ d = (a-b-a-b) = -2b \ \ \ \ \ n=22$

We know $S_n =$ $\frac{n}{2}$ $\Big[ 2a + ( n - 1) d \Big]$

$\therefore S_{22} =$ $\frac{22}{2}$ $\Big[ 2(a+b) + ( 22- 1) (-2b) \Big] = 11 [ 2a-40b] = 22a-440b$

vi) Given series $(x-y)^2, (x^2+y^2) , (x+y)^2, \ldots$

Given $a = (x-y)^2, \ \ \ \ \ d = (x^2+y^2)-(x-y)^2 = 2xy \ \ \ \ \ n=n$

We know $S_n =$ $\frac{n}{2}$ $\Big[ 2a + ( n - 1) d \Big]$

$\therefore S_{n} =$ $\frac{n}{2}$ $\Big[ 2(x-y)^2 + ( n- 1) (2xy) \Big] = n [(x-y)^2+(n-1)xy]$

vii) Given series $\frac{x-y}{x+y}, \frac{3x-2y}{x+y}, \frac{5x-3y}{x+y}, \ldots$

Given $a =$ $\frac{x-y}{x+y}$ $, \ \ \ \ \ d = \Big($ $\frac{3x-2y}{x+y}$ $-$ $\frac{x-y}{x+y}$ $\Big) =$ $\frac{2x-y}{x+y}$ $n=n$

We know $S_n =$ $\frac{n}{2}$ $\Big[ 2a + ( n - 1) d \Big]$

$\therefore S_{n} =$ $\frac{n}{2}$ $\Big[ 2 \Big($ $\frac{x-y}{x+y}$ $\Big) + ( n- 1) \Big($ $\frac{2x-y}{x+y}$ $\Big) \Big]$

$=$ $\frac{n}{2(x+y)}$ $\Big[ 2(x-y)+(n-1)(2x-y) ]$

$=$ $\frac{n}{2(x+y)}$ $[ 2nx - ny - y ]$

$=$ $\frac{n}{2(x+y)}$ $[ n(2x-y) - y ]$

$\\$

Question 2: Find the sum of the following series:

i) $2+5+8+\ldots +182$

ii) $101 +99+97 +\ldots +47$

iii) $(a-b)^2 + (a^2+b^2) + ( a+b)^2 + \ldots + [(a+b)^2+6ab]$

Answer:

i) Given series $2+5+8+\ldots +182$

Therefore $a = 2 \ \ \ \ \ d = ( 5-2) = 3 \ \ \ \ \ a_n = 182$

We know $a_n = a + (n-1)d$

$\therefore 182 = 2 + ( n-1) ( 3)$

$\Rightarrow 182 = 3n - 1$

$\Rightarrow 3n = 184$

$\Rightarrow n = 61$

We know $S_n =$ $\frac{n}{2}$ $\Big[ 2a + ( n - 1) d \Big]$

$\therefore S_{61} =$ $\frac{61}{2}$ $\Big[ 2(2) + ( 61 - 1) (3) \Big] =$ $\frac{61}{2}$ $(4+180) = 5612$

ii) Given series $101 +99+97 +\ldots +47$

Therefore $a = 101 \ \ \ \ \ d = ( 99-101) = -2 \ \ \ \ \ a_n = 47$

We know $a_n = a + (n-1)d$

$\therefore 47 = 101 + ( n-1) ( -2)$

$\Rightarrow 47 = 103 - 2n$

$\Rightarrow 2n = 56$

$\Rightarrow n = 28$

We know $S_n =$ $\frac{n}{2}$ $\Big[ 2a + ( n - 1) d \Big]$

$\therefore S_{28} =$ $\frac{28}{2}$ $\Big[ 2(101) + ( 28 - 1) (-2) \Big] = 14 [ 202-54 ] = 2072$

iii) Given series $(a-b)^2 + (a^2+b^2) + ( a+b)^2 + \ldots + [(a+b)^2+6ab]$

Therefore $a = (a-b)^2 \ \ \ \ \ d = (a^2+b^2) - (a-b)^2 = 2ab \ \ \ \ \ a_n = (a+b)^2+6ab$

We know $a_n = a + (n-1)d$

$\therefore (a+b)^2+6ab = (a-b)^2 + ( n-1) ( 2ab)$

$\Rightarrow a^2 + b^2 + 2ab + 6 ab = a^2 + b^2 - 2ab + ( n-1) ( 2ab)$

$\Rightarrow (n-1)(2ab) = 10ab$

$\Rightarrow n = 6$

We know $S_n =$ $\frac{n}{2}$ $\Big[ 2a + ( n - 1) d \Big]$

$\therefore S_{6} =$ $\frac{6}{2}$ $\Big[ 2(a-b)^2 + ( 6 - 1) (2ab) \Big]$

$= 3 [ 2a^2+2b^2 - 4ab + 10ab ] = 3 [ 2a^2 + 2b^2 + 6ab]$

$\\$

Question 3: Find the sum of first $n$ natural numbers.

Answer:

Given series $1, 2, 3, \ldots$

Therefore $a = 1 \ \ \ \ \ d = ( 2-1) = 1$

We know $S_n =$ $\frac{n}{2}$ $\Big[ 2a + ( n - 1) d \Big]$

$\therefore S_{n} =$ $\frac{n}{2}$ $\Big[ 2(1) + ( n - 1) (1) \Big] =$ $\frac{n}{2}$ $(n+1)$

$\\$

Question 4: Find the sum of all natural numbers between $1$ and $100$ , which are divisible by $2$ or $5$ .

Answer:

Sequence of natural numbers between $1$ and $100$ , which are divisible by $2$ is $2, 4, 6, \ldots, 96, 98, 100$

Therefore $a = 2 \ \ \ \ \ d = ( 4-2) = 2 \ \ \ \ \ a_n = 100$

We know $a_n = a + (n-1)d$

$\therefore 100 = 2 + ( n-1) ( 2)$

$\Rightarrow 49 = n - 1$

$\Rightarrow n = 50$

We know $S_n =$ $\frac{n}{2}$ $\Big[ 2a + ( n - 1) d \Big]$

$\therefore S_{50} =$ $\frac{50}{2}$ $\Big[ 2(2) + ( 50 - 1) (2) \Big] = 25 (4+98) = 2550$

Sequence of natural numbers between $1$ and $100$ , which are divisible by $5$ is $5, 10, 15, \ldots, 95, 100$

Therefore $a = 5 \ \ \ \ \ d = ( 10-5) = 5 \ \ \ \ \ a_n = 100$

We know $a_n = a + (n-1)d$

$\therefore 100 = 5 + ( n-1) ( 5)$

$\Rightarrow 95 = (n - 1)5$

$\Rightarrow n = 19+1$

$\Rightarrow n = 20$

We know $S_n =$ $\frac{n}{2}$ $\Big[ 2a + ( n - 1) d \Big]$

$\therefore S_{20} =$ $\frac{20}{2}$ $\Big[ 2(5) + ( 20 - 1) (5) \Big] = 10 (10+95) = 1050$

Sequence of natural numbers between $1$ and $100$ , which are divisible by $10$ is $10, 20, 30, \ldots, 90, 100$

Therefore $a = 10 \ \ \ \ \ d = ( 20-10) = 10 \ \ \ \ \ a_n = 100$

We know $a_n = a + (n-1)d$

$\therefore 100 = 10 + ( n-1) ( 10)$

$\Rightarrow 90 = (n - 1)(10)$

$\Rightarrow n = 9+1$

$\Rightarrow n = 10$

We know $S_n =$ $\frac{n}{2}$ $\Big[ 2a + ( n - 1) d \Big]$

$\therefore S_{20} =$ $\frac{10}{2}$ $\Big[ 2(10) + ( 10 - 1) (10) \Big] = 5 (20+90) = 550$

Hence, sum of all natural numbers between $1$ and $100$ , which are divisible by $2$ or $5 = 2550 + 1050 - 550 = 3050$

$\\$

Question 5: Find the sum of first $n$ odd natural numbers.

Answer:

The series would be $1, 3, 5, 7, 9, \ldots$

Therefore $a = 1 \ \ \ \ \ d = ( 3-1) = 2$

We know $S_n =$ $\frac{n}{2}$ $\Big[ 2a + ( n - 1) d \Big]$

$\therefore S_{n} =$ $\frac{n}{2}$ $\Big[ 2(1) + ( n - 1) (2) \Big] =$ $\frac{n}{2}$ $(2n) = n^2$

$\\$

Question 6: Find the sum of all odd numbers between $100$ and $200$ .

Answer:

The series is $101, 103, \ldots , 199$

Therefore $a = 101 \ \ \ \ \ d = ( 103-101) = 2 \ \ \ \ \ a_n = 199$

We know $a_n = a + (n-1)d$

$\therefore 199 = 101 + ( n-1) ( 2)$

$\Rightarrow 98 = (n - 1)(2)$

$\Rightarrow n = 49+1$

$\Rightarrow n = 50$

We know $S_n =$ $\frac{n}{2}$ $\Big[ 2a + ( n - 1) d \Big]$

$\therefore S_{50} =$ $\frac{50}{2}$ $\Big[ 2(101) + ( 50 - 1) (2) \Big] = 25 (202+98) = 7500$

$\\$

Question 7: Show that the sum of all odd integers between $1$ and $1000$ which are divisible by $3$ is $83667$ .

Answer:

The series is $3, 9, 15, \ldots , 999$

Therefore $a = 3 \ \ \ \ \ d = ( 9-3) = 6 \ \ \ \ \ a_n = 999$

We know $a_n = a + (n-1)d$

$\therefore 999 = 3 + ( n-1) ( 6)$

$\Rightarrow 996 = (n - 1)(6)$

$\Rightarrow n = 166+1$

$\Rightarrow n = 167$

We know $S_n =$ $\frac{n}{2}$ $\Big[ 2a + ( n - 1) d \Big]$

$\therefore S_{167} =$ $\frac{167}{2}$ $\Big[ 2(3) + ( 167 - 1) (6) \Big] =$ $\frac{167}{2}$ $(6+996) = 83667$

$\\$

Question 8: Find the sum of all integers between $84$ and $219$ , which are multiples of $5$ .

Answer:

The series is $85,90,95, \ldots , 710, 715$

Therefore $a = 85 \ \ \ \ \ d = ( 90-85) = 5 \ \ \ \ \ a_n = 715$

We know $a_n = a + (n-1)d$

$\therefore 715 = 85 + ( n-1) ( 5)$

$\Rightarrow 630 = (n - 1)(5)$

$\Rightarrow n = 126+1$

$\Rightarrow n = 127$

We know $S_n =$ $\frac{n}{2}$ $\Big[ 2a + ( n - 1) d \Big]$

$\therefore S_{127} =$ $\frac{127}{2}$ $\Big[ 2(85) + ( 127 - 1) (5) \Big] =$ $\frac{127}{2}$ $(170+630) = 50800$

$\\$

Question 9: Find the sum of all integers between $50$ and $500$ which are divisible by $7$ .

Answer:

The series is $56, 63, \ldots , 497$

Therefore $a = 56 \ \ \ \ \ d = ( 63-56) = 7 \ \ \ \ \ a_n = 497$

We know $a_n = a + (n-1)d$

$\therefore 497 = 56 + ( n-1) ( 7)$

$\Rightarrow 441 = (n - 1)(7)$

$\Rightarrow 7n = 448$

$\Rightarrow n = 64$

We know $S_n =$ $\frac{n}{2}$ $\Big[ 2a + ( n - 1) d \Big]$

$\therefore S_{64} =$ $\frac{64}{2}$ $\Big[ 2(56) + ( 64 - 1) (7) \Big] = 32 (112+441) = 17696$

$\\$

Question 10: Find the sum of all even integers between $101$ and $999$ .

Answer:

The series is $102, 104, \ldots , 998$

Therefore $a = 102 \ \ \ \ \ d = ( 104-102) = 2 \ \ \ \ \ a_n = 998$

We know $a_n = a + (n-1)d$

$\therefore 998 = 102 + ( n-1) ( 2)$

$\Rightarrow 896 = (n - 1)(2)$

$\Rightarrow n = 448+1$

$\Rightarrow n = 449$

We know $S_n =$ $\frac{n}{2}$ $\Big[ 2a + ( n - 1) d \Big]$

$\therefore S_{449} =$ $\frac{449}{2}$ $\Big[ 2(102) + ( 449 - 1) (2) \Big] =$ $\frac{449}{2}$ $[ 204+896] = 246950$

$\\$

Question 11: Find the sum of all integers between $100$ and $550$ , which are divisible by $9$ .

Answer:

The series is $108, 117, \ldots , 549$

Therefore $a = 108 \ \ \ \ \ d = ( 117-108) = 9 \ \ \ \ \ a_n = 549$

We know $a_n = a + (n-1)d$

$\therefore 549 = 108 + ( n-1) ( 9)$

$\Rightarrow 441 = (n - 1)(9)$

$\Rightarrow n = 49+1$

$\Rightarrow n = 50$

We know $S_n =$ $\frac{n}{2}$ $\Big[ 2a + ( n - 1) d \Big]$

$\therefore S_{50} =$ $\frac{50}{2}$ $\Big[ 2(108) + ( 50 - 1) (9) \Big] = 25 [ 216+441] = 16425$

$\\$

Question 12: Find the sum of the series: $3+5+7+6+9+12+9 +13+17 + \ldots$ to $3n$ terms.

Answer:

Given sequence $3, 5, 7, 6, 9, 12, 9, 13, 17, \ldots$

This can we written as $[ 3, 6, 9, \ldots ] , [ 5, 9, 13, \ldots ] \ \ \& \ \ [ 7, 12, 17, \ldots ]$

For $[ 3, 6, 9, \ldots ], \ \ \ a = 3, \ \ \ \ d = 6-3 = 3$

For $[ 5, 9, 13, \ldots ], \ \ \ \ a = 5, \ \ \ \ d = 9-5 = 4$

For $[ 7, 12, 17, \ldots ], \ \ \ \ a = 7, \ \ \ \ d = 12-7 = 5$

$\therefore$ Sum $=$ $\frac{n}{2}$ $[ 2(3) +(n-1)(3) ] +$ $\frac{n}{2}$ $[ 2(5) +(n-1)(4) ] +$ $\frac{n}{2}$ $[ 2(7) +(n-1)(5) ]$

$=$ $\frac{n}{2}$ $[ 6 + 3(n-1) + 10 + 4( n-1) + 14 + 5( n-1) ]$

$=$ $\frac{n}{2}$ $[ 30 + 12( n-1) ]$

$= n [ 15 + 6(n-1) ]$

$= n [ 6n + 9 ]$

$= 3n [2n+3]$

$\\$

Question 13: Find the sum of all those integers between $100$ and $800$ each of which on division leaves the remainder $7$

Answer:

The number would be of the form $N = 16q + 7$

The series is $103, 119, \ldots , 791$

Therefore $a = 103 \ \ \ \ \ d = ( 119-103) = 16 \ \ \ \ \ a_n = 791$

We know $a_n = a + (n-1)d$

$\therefore 791 = 103 + ( n-1) ( 16)$

$\Rightarrow 688 = (n - 1)(16)$

$\Rightarrow 16n = 704$

$\Rightarrow n = 44$

We know $S_n =$ $\frac{n}{2}$ $\Big[ 2a + ( n - 1) d \Big]$

$\therefore S_{44} =$ $\frac{44}{2}$ $\Big[ 2(103) + ( 44 - 1) (16) \Big] = 22 [ 206+688] = 19668$

$\\$

Question 14: Solve: (i) $25 + 22 + 19 + 16 +... + x = 115$ (ii) $1+4+7 +10+ \ldots x=590$ .

Answer:

(i) $25 + 22 + 19 + 16 +... + x = 115$

$\therefore S_n = 115$

$a = 25 \ \ \ \ \ d = ( 22-25) = -3$ Let the number of terms be $n$

We know $S_n =$ $\frac{n}{2}$ $\Big[ 2a + ( n - 1) d \Big]$

$\Rightarrow 115 =$ $\frac{n}{2}$ $\Big[ 2(25) + ( n-1)(-3) \Big]$

$\Rightarrow 230 = n [ 50-3(n-1) ]$

$\Rightarrow 230 = 50 n - 3n^2 + 3n$

$\Rightarrow 3n^2 - 53n + 230 = 0$

$\Rightarrow (3n-23)(n-10)= 0$

$\Rightarrow n =$ $\frac{23}{3}$ $\ or \ n = 10$

Since $n \neq$ $\frac{23}{3}$ $\therefore n = 10$

We know $a_n = a + (n-1)d$

$\therefore x = 25 + ( 10 - 1) ( -3)$

$\Rightarrow x = 25 - 27 = -2$

(ii) $1+4+7 +10+ \ldots x=590$

$\therefore S_n = 590$

$a = 1 \ \ \ \ \ d = ( 4-1) = 3$ Let the number of terms be $n$

We know $S_n =$ $\frac{n}{2}$ $\Big[ 2a + ( n - 1) d \Big]$

$\Rightarrow 590 =$ $\frac{n}{2}$ $\Big[ 2(1) + ( n-1)(3) \Big]$

$\Rightarrow 590 =$ $\frac{n}{2}$ $\Big[ 2 + 3n - 3 \Big]$

$\Rightarrow 590 =$ $\frac{n}{2}$ $\Big[ 3n-1 \Big]$

$3n^2 - n - 1180 = 0$

$(3n-59)(n-20) = 0$

$n =$ $\frac{59}{3}$ or $n = 20$

Since $n \neq$ $\frac{59}{3}$ $\ \ \ \ \therefore n = 20$

We know $a_n = a + (n-1)d$

$\therefore x = 1 + ( 20 - 1) ( 3)$

$\Rightarrow x = 1+57 = 58$

$\\$

Question 15: Find the $r^{th}$ term of an A.P., the sum of whose first $n$ terms is $3n^2 - 2n$ .

Answer:

Let the first term $= a$ and the common difference $= d$

Given $S_n = 3n^2 + 2n$

$S_1 = 3(1)^2 + 2(1) = 5$

$S_2 = 3(2)^2 + 2(2) = 12 + 4 = 16$

$\therefore a + ( a+d) = 16$

$\Rightarrow 5 + 5 + d = 16$

$\Rightarrow d = 6$

$\therefore a_r = a + ( r-1) d$

$a_r = 5 + ( r-1) (6)$

$a_r = 6r-1$

$\\$

Question 16: How many terms are there in the A.P. whose first and fifth terms are $-14$ and $2$ respective and the sum of the terms is $40$ ?

Answer:

Let the first term $= a$ and the common difference $= d$

Given $a = -14$

$a+ ( 4-1) (d) = 2 \Rightarrow -14 + 3( d) = 2 \Rightarrow d = 4$

We know $S_n =$ $\frac{n}{2}$ $\Big[ 2a + ( n - 1) d \Big]$

$\therefore 40 =$ $\frac{n}{2}$ $\Big[ -28 + ( n - 1) (4) \Big]$

$\Rightarrow 4n^2 - 32 n - 80 = 0$

$\Rightarrow n^2 - 8 n - 20 = 0$

$\Rightarrow (n-10)(n+2) = 0$

$\Rightarrow n = 10$ or $n = -2$ (this is not possible)

Therefore there are $10$ terms in the A.P.

$\\$

Question 17: The sum of first $7$ terms of an A.P. is $10$ and that of next $7$ terms is $17$ . Find the progression.

Answer:

Given $S_7 = 10$

$\Rightarrow$ $\frac{7}{2}$ $[ 2a+6d] = 10$

$\Rightarrow 2a + 6d =$ $\frac{20}{7}$

$\Rightarrow a + 3d =$ $\frac{10}{7}$ … … … … … i)

Also $S_{14}- S_7 = 17$

$\Rightarrow$ $\frac{14}{2}$ $[ 2a + 13d] -$ $\frac{7}{2}$ $[ 2a + 6d] = 17$

$\Rightarrow 28a + 182d - 14a - 42d = 34$

$\Rightarrow 14a + 140 d = 34$

$\Rightarrow 7a + 70 d = 17$ … … … … … ii)

Solving i) and ii)

$7 [$ $\frac{10}{7}$ $- 3d ] + 70d = 17$

$\Rightarrow 10 - 21d + 70d = 17$

$\Rightarrow 49 d = 7$

$\Rightarrow d = 7$

$\therefore a =$ $\frac{10}{7}$ $- 3($ $\frac{1}{7}$ $) =$ $\frac{7}{7}$ $= 1$

Therefore the series is $1,$ $\frac{8}{7}$ $,$ $\frac{9}{7}$ $,$ $\frac{10}{7}$ $, \ldots$

$\\$

Question 18: The third term of an A .P. is $7$ and the seventh term exceeds three times the third term by $2$ . Find the first term, the common difference and the sum of first $20$ term.

Answer:

Given $a_3 = 7$

We know $a_n = n + (n-1) d$

$\therefore 7 = a + ( 3-1) d$

$\Rightarrow a + 2d = 7$ … … … … … i)

Also $a_7 - 3a_3 = 2$

$\Rightarrow a_7 - 3(7) = 2$

$\Rightarrow a_7 = 23$

$\Rightarrow a + ( 7-1) d = 23$

$\Rightarrow a + 6d = 23$ … … … … … ii)

Solving i) and ii) we get

$7 - 2d = 23-6d$

$\Rightarrow 4d = 16$

$\Rightarrow d = 4$

$\therefore a = 7 - 2( 4) = -1$

We know $S_n =$ $\frac{n}{2}$ $[ 2(a) + ( n-1)(d) ]$

$\therefore S_{20}=$ $\frac{20}{2}$ $[ 2(-1) + ( 20 -1) ( 4) ] = 10 [ -2 + 76] = 740$

$\therefore a = -1 , d = 4,$ and $S_{20} = 740$

$\\$

Question 19: The first term of an A.P. is $2$ and the last term is $50$ . The sum of all these terms is $442$ . Find the common difference.

Answer:

Given $a =2 , \ \ \ \ l = 50, \ \ \ \ \ S_n = 442$

$l = a + ( n-1)d$

$\Rightarrow 50 = 2 + ( n-1) d$

$\Rightarrow (n-1) d = 48$ … … … … … i)

We know $S_n =$ $\frac{n}{2}$ $[ 2(a) + ( n-1)(d) ]$

$\Rightarrow 442 =$ $\frac{n}{2}$ $[ a + l]$

$\Rightarrow 442 =$ $\frac{n}{2}$ $[ 2+50]$

$\Rightarrow n = 17$

$\therefore d =$ $\frac{48}{17-1}$ $=$ $\frac{48}{16}$ $= 3$

Therefore common difference is $3$ .

$\\$

Question 20: The number of terms of an A.P. is even; the sum of odd terms is $24$ , of the even terms is $30$ , and the last term exceeds the first by $10$ $\frac{1}{2}$ , find the number of term and the series.

Answer:

Let the total number of terms be $2n$

Let $a_1 + a_3 + \ldots + a_{2n-1} = 24$ … … … … … i)

$a_2 + a_4 + \ldots + a_{2n} = 30$ … … … … … ii)

Subtracting i) from ii) we get

$( a_2 - a_1) + ( a_2 - a_1) + \ldots + ( a_{2n} - a{2n-1}) = 6$

$\Rightarrow d + d + \ldots + d = 6$

$\Rightarrow nd = 6$

Also $a_{2n} = a_1 +$ $\frac{21}{2}$

$\Rightarrow a+ (2n-1) d = a +$ $\frac{21}{2}$

$\Rightarrow (2n-1) d =$ $\frac{21}{2}$

$\Rightarrow 2nd - d =$ $\frac{21}{2}$

$\Rightarrow 2 \times 6 - d =$ $\frac{21}{2}$

$\Rightarrow d = 12 -$ $\frac{21}{2}$ $=$ $\frac{3}{2}$

$\therefore n =$ $\frac{6}{\frac{3}{2}}$ $= 4$

Therefore there are $8$ terms in the series.

Now $a_2 + a_4 + \ldots + a_{2n} = 30$

$\Rightarrow (a+d) + ( a+3d) + \ldots + [ a + (2n-1) d] = 30$

$\Rightarrow$ $\frac{n}{2}$ $\Big[ (a+d) + a + (2n-1) d \Big] = 30$

$\Rightarrow$ $\frac{4}{2}$ $\Big[ (a+$ $\frac{3}{2}$ $) + a + (2 \times 4 - 1)$ $\frac{3}{2}$ $\Big] = 30$

$\Rightarrow 2 \Big[ 2a +$ $\frac{3}{2}$ $+ 7 \cdot$ $\frac{3}{2}$ $\Big] = 30$

$\Rightarrow 2 [ 2a + 12] = 30$

$\Rightarrow 2a = 15 - 12$

$\Rightarrow a =$ $\frac{3}{2}$

Therefore series is $1$ $\frac{3}{2}$ $, 3, 4$ $\frac{1}{2}$ $, \ldots$

$\\$

Question 21: If $S_n =n^2 p$ and, $S_m = m^2 p, m \neq n$ , in an A.P., prove that $S_p = p^3$ .

Answer:

Given $S_n = n^2p$

$\Rightarrow$ $\frac{n}{2}$ $[ 2a + (n-1) d ] = n^2p$

$\Rightarrow 2a + ( n-1) d = 2n p$ … … … … … i)

Also $S_m = m^2 p$

$\Rightarrow$ $\frac{m}{2}$ $[ 2a +(m-1) d ] = m^2 p$

$\Rightarrow 2a + ( m-1) d= 2mp$ … … … … … ii)

Solving i) and ii) we get

$2np - (n-1) d = 2mp - ( m-1) d$

$\Rightarrow 2p( n-m) = ( n-1-m+1)d$

$\Rightarrow 2p(n-m) = (n-m) d$

$\Rightarrow 2p = d$ … … … … … iii)

Substituting back in i) we get

$2a = 2np - ( n-1) (2p)$

$\Rightarrow 2a = 2np - 2np + 2p$

$\Rightarrow a = p$

$\therefore S_p =$ $\frac{p}{2}$ $[ 2a + ( p-1) d ]$

$\Rightarrow S_p =$ $\frac{p}{2}$ $[ 2p + ( p-1) (2p) ]$

$\Rightarrow S_p =$ $\frac{p}{2}$ $[ 2p + 2p^2 - 2p) ]$

$\Rightarrow S_p = p^3$

$\\$

Question 22: If $12^{th}$ term of an A.P. is $-13$ and the sum of the first four terms is $24$ , what is the sum of first $10$ terms ?

Answer:

Given $a_{12} = -13$ and $S_4 = 24$

We know $a_n = a + ( n-1) d$

$\Rightarrow -13 = a + ( 12 - 1) d$

$\Rightarrow a + 11 d = - 13$ … … … … … i)

We know $S_n =$ $\frac{n}{2}$ $[ 2a + ( n-1) d]$

$\Rightarrow 24 =$ $\frac{4}{2}$ $[ 2( a) + ( 4 -1 ) d ]$

$\Rightarrow 2a + 3d = 12$ … … … … … ii)

Solving i) and ii) we get

$\Rightarrow 2 ( - 13 - 11d) + 3d = 12$

$\Rightarrow -26 - 22d + 3d = 12$

$\Rightarrow -19d = 38$

$\Rightarrow d = - 2$

$\therefore a = - 13 - 11 ( -2) = -13 + 22 = 9$

$\therefore S_{10} =$ $\frac{10}{2}$ $[ 2 ( 9) + ( 10-1)( -2) ] = 5 [ 18-18] = 0$

$\\$

Question 23: It the $5^{th}$ and $12^{th}$ terms of an A.P. are $30$ and $65$ respectively, what is the sum of the first $20$ terms ?

Answer:

Given $a_{5} = 30$ and $a_{12} = 65$

We know $a_n = a + ( n-1) d$

$\Rightarrow 30 = a + ( 5 - 1) d$

$\Rightarrow a + 4 d = 30$ … … … … … i)

Similarly, $\Rightarrow 65 = a + ( 12 - 1) d$

$\Rightarrow a + 11 d = 65$ … … … … … ii)

Solving i) and ii) we get

$\Rightarrow 30-4d = 65-11d$

$\Rightarrow 7d = 35$

$\Rightarrow d = 5$

$\therefore a =30-4(5) = 10$

$\therefore S_{20} =$ $\frac{20}{2}$ $[ 2 ( 10) + ( 20-1)( 5) ] = 10 [ 20+95] = 1150$

$\\$

Question 24: Find the sum of $n$ terms of the A.P. whose $k^{th}$ terms is $5k + 1$ .

Answer:

Given $a_k = 5k + 1$

For $k = 1, a_1 = 5(1) + 1 = 6$

For $k = 2, a_2 = 5(2) + 1 = 11$

$\therefore d = a_2 - a_1 = 11 - 6 = 5$

$\therefore S_n =$ $\frac{n}{2}$ $[ 2(6) + ( n-1) (5) ]$

$=$ $\frac{n}{2}$ $[12+5n-5]$

$=$ $\frac{n}{2}$ $(5n+7)$

$\\$

Question 25: Find the sum of all two digit numbers which when divided by $4$ , yields $1$ as remainder.

Answer:

The number would be of the form $N = 4q + 1$

Therefore the series would be $13, 17, \ldots, 97$ for a two digit number

$\therefore a = 13, \ \ \ \ \ d = (17-13) = 4 \ \ \ \ \ a_n = 97$

We know $a_n = a + ( n-1) d$

$\Rightarrow 97 = 13 + ( n-1) ( 4)$

$\Rightarrow 4n = 88$

$\Rightarrow n = 22$

We know $S_n =$ $\frac{n}{2}$ $[ 2a + ( n-1) d]$

$\therefore$ $S_{22} =$ $\frac{22}{2}$ $[ 2(13) + ( 22-1) (4)] = 11 [ 26+84] = 1210$

$\\$

Question 26: If the sum of a certain number of terms of the A.P. $25,22,19, \ldots$ is $116$ . Find the last term.

Answer:

Given series $25,22,19, \ldots$

$\therefore a = 25, \ \ \ \ \ d = (22-25) = -2 \ \ \ \ \ S_n = 116$

We know $S_n =$ $\frac{n}{2}$ $[ 2a + ( n-1) d]$

$\therefore$ $116 =$ $\frac{n}{2}$ $[ 2(25) + ( n-1) (-3)]$

$232 = n [ 50 - 3n + 4 ]$

$3n^2 - 54n + 232 = 0$

$(3n-29)(n-8) = 0$

$\therefore n = \frac{29}{3} \ or \ n = 8$

Since $n \neq \frac{29}{3} \therefore n = 8$

$\therefore a_8 = 25 + ( 8-1) (-3) = 25 - 21 = 4$

$\\$

Question 27: Find the sum of odd integers from $1$ to $2001$ .

Answer:

The series will be $1, 3, 5, \ldots , 2001$

$\therefore a = 1 \ \ \ \ \ d = (3-1) = 2 \ \ \ \ \ a_n = 2001$

We know $a_n = a + ( n-1) d$

$\therefore 2001 = 1 + ( n - 1) (2)$

$\Rightarrow 2000 = ( n-1) ( 2)$

$\Rightarrow n = 1001$

We know $S_n =$ $\frac{n}{2}$ $\Big[ 2(a) + (n-1) d \Big]$

$\therefore S_{1001} =$ $\frac{1001}{2}$ $\Big [ 2(10 + ( 1001 - 1)(2) \Big] =$ $\frac{1001}{2}$ $\Big [ 2 + 20000 \Big] = 1002001$

$\\$

Question 28: How many terms of the A.P. $- 6,$ $\frac{-11}{5}$ $, -5, \ldots$ are needed to give the sum $- 251$ .

Answer:

Given series $- 6,$ $\frac{-11}{5}$ $, -5, \ldots$

$\therefore a = -6 \ \ \ \ \ d = \Big[$ $\frac{-11}{5}$ $- (-6) \Big] =$ $\frac{1}{2}$ $\ \ \ \ \ S_n = -25$

We know $S_n =$ $\frac{n}{2}$ $\Big[ 2(a) + (n-1) d \Big]$

$\therefore -25 =$ $\frac{n}{2}$ $\Big [ 2( -6) + ( n - 1) \Big($ $\frac{1}{2}$ $\Big) \Big]$

$- 25 =$ $\frac{n}{2}$ $\Big [ - 12 +$ $\frac{n}{2}$ $-$ $\frac{1}{2}$ $\Big ]$

$-50 = n \Big[$ $\frac{n}{2}$ $-$ $\frac{25}{2}$ $\Big ]$

$-100 = n^2 - 25 n$

$n^2 - 25 n + 100 = 0$

$(n-20)(n-5) = 0$

$n = 20$ or $n = 5$

$\\$

Question 29: In an A.P. the first term is $2$ and the sum of the first five terms is one fourth of the next five terms. Show that $20^{th}$ term is $-112$ .

Answer:

Given $a = 2 , \ \ \ \ S_5 =$ $\frac{1}{4}$ $\Big(S_{10}-S_5 \Big)$

We know $S_n =$ $\frac{n}{2}$ $\Big[ 2(a) + (n-1) d \Big]$

$\therefore S_5 =$ $\frac{5}{2}$ $\Big [ 2(2) + (5-1) d \Big] = 5 \Big[ 2 + 2d \Big]$ … … … … … i)

Similarly, $S_{10} =$ $\frac{10}{2}$ $\Big [ 2(2) + ( 10 - 1) d \Big] = 5 \Big[ 4 + 9d \Big]$ … … … … … ii)

Substituting in i)

$5 [ 2 + 2d] =$ $\frac{1}{4}$ $\Big [ 5( 4 + 9d) - 5 ( 2+2d) \Big]$

$8+8d = 4 + 9d - 2 - 2d$

$8d + 8 = 7d + 2$

$d = - 6$

We know $a_n = a + ( n-1) d$

$\therefore a_{20} = 2 + ( 20-1) (-6) = - 112$

$\\$

Question 30: If $S_1$ be the sum of $(2n + 1)$ terms of an A.P. and $S_2$ be the sum of its odd terms, then prove that: $S_1:S_2 = (2n+1):(n+1)$ .

Answer:

Let the A.P. be $a, a+d, a+2d, \ldots$

$\therefore S_1 =$ $\frac{2n+1}{2}$ $\Big[ 2(a) + ( 2n+1-1) d \Big]$

$\Rightarrow S_1 = (2n+1)(a+nd)$ … … … … … i)

Similarly, $S_2 =$ $\frac{n+1}{2}$ $\Big[ 2(a) + ( n+ 1 - 1)(2d) \Big]$

$\Rightarrow S_2 = ( n+1) ( a + nd)$ … … … … … ii)

$\therefore$ $\frac{S_1}{S_2}$ $=$ $\frac{ (2n+1)(a+nd)}{( n+1) ( a + nd)}$ $=$ $\frac{2n+1}{n+1}$ . Hence proved.

$\\$

Question 31: Find an A.P. in which the sum of any number of terms is always three times the squared number of these terms.

Answer:

Given $S_n = 3n^2$

For $n = 1, S_1 = 3(1)^2 = 3$

For $n = 2 , S_2 = 3(2)^2 = 12$

For $n = 3, S_3 = (3)^2 = 27$

$\therefore a_1= 3$

$a_2 = 12 - 3 = 9$

$a_3 = 27 - 12 = 15$

$\therefore d = 9 - 3 = 6$

$\therefore$ series is $3, 9, 15, \ldots$

$\\$

Question 32: If the sum of n terms of an A.P. is $nP +$ $\frac{1}{2}$ $n(n-1)Q$ , where $P$ and $Q$ are constants, find the common difference.

Answer:

Given $S_n = nP +$ $\frac{1}{2}$ $n(n-1)Q$

For $n = 1, S_1 = (1) P +$ $\frac{1}{2}$ $(1) ( 1-1) Q = P$

For $n = 2, S_2 = (2) P +$ $\frac{1}{2}$ $(2) (2-1) Q = 2P+Q$

$a_1 = P$

$a_2 = S_2 - S_1 = 2P + Q - P = P+Q$

$\therefore d = a_2 - a_1 = P+Q - P = Q$

$\\$

Question 33: The sums of $n$ terms of two arithmetic progressions are in the ratio $5n + 4 :9n + 6$ . Find the ratio of their $18^{th}$ terms.

Answer:

Let for first A.P. $a = a_1$ and $d= d_1$

Let for second A.P. $a = a_2$ and $d= d_2$

Given $\frac{S_{n1}}{S_{n2}}$ $=$ $\frac{5n+4}{9n+6}$

$\Rightarrow$ $\frac{5n+4}{9n+6}$ $=$ $\frac{\frac{n}{2} [ 2a_1 + ( n - 1)d_1]}{\frac{n}{2} [ 2a_2 + ( n - 1)d_2]}$

$\Rightarrow$ $\frac{5n+4}{9n+6}$ $=$ $\frac{2a_1 + ( n - 1)d_1}{2a_2 + ( n - 1)d_2}$

If you put $n = 35$ we get

$\Rightarrow$ $\frac{5(35)+4}{9(35)+6}$ $=$ $\frac{2a_1 + ( 35 - 1)d_1}{2a_2 + ( 35 - 1)d_2}$

$\Rightarrow$ $\frac{179}{321}$ $=$ $\frac{2[ a_1 + 17d_1]}{2[ a_2 + 17d_2]}$

$\Rightarrow$ $\frac{179}{321}$ $=$ $\frac{ a_1 + (18-1)d_1}{ a_2 + (18-1)d_2}$

$\Rightarrow$ $\frac{179}{321}$ $=$ $\frac{18^{th} \text{term of first A.P.}}{18^{th} \text{term of second A.P.}}$

$\\$

Question 34: The sums of first $n$ terms of two A.P.’s are in the ratio $(7n + 2) : (n + 4)$ . Find the ratio of their $5^{th}$ terms.

Answer:

Let for first A.P. $a = a_1$ and $d= d_1$

Let for second A.P. $a = a_2$ and $d= d_2$

Given $\frac{S_{n1}}{S_{n2}}$ $=$ $\frac{7n+2}{n+2}$

$\Rightarrow$ $\frac{7n+2}{n+2}$ $=$ $\frac{\frac{n}{2} [ 2a_1 + ( n - 1)d_1]}{\frac{n}{2} [ 2a_2 + ( n - 1)d_2]}$

$\Rightarrow$ $\frac{7n+2}{n+2}$ $=$ $\frac{2a_1 + ( n - 1)d_1}{2a_2 + ( n - 1)d_2}$

If you put $n = 9$ we get

$\Rightarrow$ $\frac{7(9)+2}{9+2}$ $=$ $\frac{2a_1 + ( 9 - 1)d_1}{2a_2 + ( 9 - 1)d_2}$

$\Rightarrow$ $\frac{65}{13}$ $=$ $\frac{2[ a_1 + 4d_1]}{2[ a_2 + 4d_2]}$

$\Rightarrow$ $\frac{65}{13}$ $=$ $\frac{ a_1 + (5-1)d_1}{ a_2 + (5-1)d_2}$

$\Rightarrow$ $\frac{65}{13}$ $=$ $\frac{5^{th} \text{term of first A.P.}}{5^{th} \text{term of second A.P.}}$

ICSE / ISC / CBSE Mathematics Portal for K12 Students

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Class 11: Arithmetic Progressions – Exercise 19.4

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