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:

$\displaystyle \text{i) } 50,46, 42, \ldots \text{ to } 10 \text{ terms }$

$\displaystyle \text{ii) } 1,3, 5,7, \ldots \text{ to } 12 \text{ terms }$

$\displaystyle \text{iii) } 3, 9/2, 6, 15/2, \ldots \text{ to } 25 \text{ terms }$

$\displaystyle \text{iv) } 41, 36, 31, \ldots \text{ to } 12 \text{ terms }$

$\displaystyle \text{v) } a + b, a -b, a - 3b, \ldots \text{ to } 22 \text{ terms }$

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

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

Answer:

$\displaystyle \text{i) } \text{Given series } 50,46, 42, \ldots$

$\displaystyle \text{Given } a = 50, \ \ \ \ \ d = (46-50) = -4 \ \ \ \ \ n=10$

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

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

$\displaystyle \text{ii) } \text{Given series } 1,3, 5,7, \ldots$

$\displaystyle \text{Given } a = 1, \ \ \ \ \ d = (3-1) = 2 \ \ \ \ \ n=12$

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

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

$\displaystyle \text{iii) } \text{Given series } 3, 9/2, 6, 15/2, \ldots$

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

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

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

$\displaystyle \text{iv) } \text{Given series } 41, 36, 31, \ldots$

$\displaystyle \text{Given } a = 41, \ \ \ \ \ d = (36-41) = -5 \ \ \ \ \ n=12$

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

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

$\displaystyle \text{v) } \text{Given series } a + b, a -b, a - 3b, \ldots$

$\displaystyle \text{Given } a = a+b, \ \ \ \ \ d = (a-b-a-b) = -2b \ \ \ \ \ n=22$

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

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

$\displaystyle \text{vi) } \text{Given series } (x-y)^2, (x^2+y^2) , (x+y)^2, \ldots$

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

Question 2: Find the sum of the following series:

$\displaystyle \text{i) } 2+5+8+\ldots +182$

$\displaystyle \text{ii) } 101 +99+97 +\ldots +47$

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

Answer:

$\displaystyle \text{i) } \text{Given series } 2+5+8+\ldots +182$

$\displaystyle \text{Therefore } a = 2 \ \ \ \ \ d = ( 5-2) = 3 \ \ \ \ \ a_n = 182$

$\displaystyle \text{We know } a_n = a + (n-1)d$

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

$\displaystyle \Rightarrow 182 = 3n - 1$

$\displaystyle \Rightarrow 3n = 184$

$\displaystyle \Rightarrow n = 61$

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

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

$\displaystyle \text{ii) } \text{Given series } 101 +99+97 +\ldots +47$

$\displaystyle \text{Therefore } a = 101 \ \ \ \ \ d = ( 99-101) = -2 \ \ \ \ \ a_n = 47$

$\displaystyle \text{We know } a_n = a + (n-1)d$

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

$\displaystyle \Rightarrow 47 = 103 - 2n$

$\displaystyle \Rightarrow 2n = 56$

$\displaystyle \Rightarrow n = 28$

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

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

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

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

$\displaystyle \text{We know } a_n = a + (n-1)d$

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

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

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

$\displaystyle \Rightarrow n = 6$

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

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

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

$\displaystyle \\$

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

Answer:

$\displaystyle \text{Given series } 1, 2, 3, \ldots$

$\displaystyle \text{Therefore } a = 1 \ \ \ \ \ d = ( 2-1) = 1$

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

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

$\displaystyle \\$

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

Answer:

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

$\displaystyle \text{Therefore } a = 2 \ \ \ \ \ d = ( 4-2) = 2 \ \ \ \ \ a_n = 100$

$\displaystyle \text{We know } a_n = a + (n-1)d$

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

$\displaystyle \Rightarrow 49 = n - 1$

$\displaystyle \Rightarrow n = 50$

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

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

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

$\displaystyle \text{Therefore } a = 5 \ \ \ \ \ d = ( 10-5) = 5 \ \ \ \ \ a_n = 100$

$\displaystyle \text{We know } a_n = a + (n-1)d$

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

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

$\displaystyle \Rightarrow n = 19+1$

$\displaystyle \Rightarrow n = 20$

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

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

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

$\displaystyle \text{Therefore } a = 10 \ \ \ \ \ d = ( 20-10) = 10 \ \ \ \ \ a_n = 100$

$\displaystyle \text{We know } a_n = a + (n-1)d$

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

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

$\displaystyle \Rightarrow n = 9+1$

$\displaystyle \Rightarrow n = 10$

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

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

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

$\displaystyle \\$

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

Answer:

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

$\displaystyle \text{Therefore } a = 1 \ \ \ \ \ d = ( 3-1) = 2$

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

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

$\displaystyle \\$

Question 6: Find the sum of all odd numbers between $\displaystyle 100 \text{ and } 200 .$

Answer:

$\displaystyle \text{The series is } 101, 103, \ldots , 199$

$\displaystyle \text{Therefore } a = 101 \ \ \ \ \ d = ( 103-101) = 2 \ \ \ \ \ a_n = 199$

$\displaystyle \text{We know } a_n = a + (n-1)d$

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

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

$\displaystyle \Rightarrow n = 49+1$

$\displaystyle \Rightarrow n = 50$

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

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

$\displaystyle \\$

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

Answer:

$\displaystyle \text{The series is } 3, 9, 15, \ldots , 999$

$\displaystyle \text{Therefore } a = 3 \ \ \ \ \ d = ( 9-3) = 6 \ \ \ \ \ a_n = 999$

$\displaystyle \text{We know } a_n = a + (n-1)d$

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

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

$\displaystyle \Rightarrow n = 166+1$

$\displaystyle \Rightarrow n = 167$

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

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

$\displaystyle \\$

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

Answer:

$\displaystyle \text{The series is } 85,90,95, \ldots , 710, 715$

$\displaystyle \text{Therefore } a = 85 \ \ \ \ \ d = ( 90-85) = 5 \ \ \ \ \ a_n = 715$

$\displaystyle \text{We know } a_n = a + (n-1)d$

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

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

$\displaystyle \Rightarrow n = 126+1$

$\displaystyle \Rightarrow n = 127$

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

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

$\displaystyle \\$

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

Answer:

$\displaystyle \text{The series is } 56, 63, \ldots , 497$

$\displaystyle \text{Therefore } a = 56 \ \ \ \ \ d = ( 63-56) = 7 \ \ \ \ \ a_n = 497$

$\displaystyle \text{We know } a_n = a + (n-1)d$

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

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

$\displaystyle \Rightarrow 7n = 448$

$\displaystyle \Rightarrow n = 64$

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

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

$\displaystyle \\$

Question 10: Find the sum of all even integers between $\displaystyle 101 \text{ and } 999 .$

Answer:

$\displaystyle \text{The series is } 102, 104, \ldots , 998$

$\displaystyle \text{Therefore } a = 102 \ \ \ \ \ d = ( 104-102) = 2 \ \ \ \ \ a_n = 998$

$\displaystyle \text{We know } a_n = a + (n-1)d$

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

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

$\displaystyle \Rightarrow n = 448+1$

$\displaystyle \Rightarrow n = 449$

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

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

$\displaystyle \\$

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

Answer:

$\displaystyle \text{The series is } 108, 117, \ldots , 549$

$\displaystyle \text{Therefore } a = 108 \ \ \ \ \ d = ( 117-108) = 9 \ \ \ \ \ a_n = 549$

$\displaystyle \text{We know } a_n = a + (n-1)d$

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

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

$\displaystyle \Rightarrow n = 49+1$

$\displaystyle \Rightarrow n = 50$

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

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

$\displaystyle \\$

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

Answer:

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

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

$\displaystyle \text{For } [ 3, 6, 9, \ldots ], \ \ \ a = 3, \ \ \ \ d = 6-3 = 3$

$\displaystyle \text{For } [ 5, 9, 13, \ldots ], \ \ \ \ a = 5, \ \ \ \ d = 9-5 = 4$

$\displaystyle \text{For } [ 7, 12, 17, \ldots ], \ \ \ \ a = 7, \ \ \ \ d = 12-7 = 5$

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

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

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

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

$\displaystyle = n [ 6n + 9 ]$

$\displaystyle = 3n [2n+3]$

$\displaystyle \\$

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

Answer:

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

$\displaystyle \text{The series is } 103, 119, \ldots , 791$

$\displaystyle \text{Therefore } a = 103 \ \ \ \ \ d = ( 119-103) = 16 \ \ \ \ \ a_n = 791$

$\displaystyle \text{We know } a_n = a + (n-1)d$

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

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

$\displaystyle \Rightarrow 16n = 704$

$\displaystyle \Rightarrow n = 44$

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

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

$\displaystyle \\$

Question 14: Solve: $\displaystyle \text{i) } 25 + 22 + 19 + 16 +... + x = 115 \hspace{1.0cm} \displaystyle \text{ii) } 1+4+7 +10+ \ldots x=590 .$

Answer:

$\displaystyle \text{i) } 25 + 22 + 19 + 16 +... + x = 115$

$\displaystyle \therefore S_n = 115$

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

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

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

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

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

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

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

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

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

$\displaystyle \text{We know } a_n = a + (n-1)d$

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

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

$\displaystyle \text{ii) } 1+4+7 +10+ \ldots x=590$

$\displaystyle \therefore S_n = 590$

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

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

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

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

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

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

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

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

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

$\displaystyle \text{We know } a_n = a + (n-1)d$

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

$\displaystyle \Rightarrow x = 1+57 = 58$

$\displaystyle \\$

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

Answer:

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

$\displaystyle \text{Given } S_n = 3n^2 + 2n$

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

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

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

$\displaystyle \Rightarrow 5 + 5 + d = 16$

$\displaystyle \Rightarrow d = 6$

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

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

$\displaystyle a_r = 6r-1$

$\displaystyle \\$

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

Answer:

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

$\displaystyle \text{Given } a = -14$

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

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

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

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

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

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

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

Therefore there are $\displaystyle 10 \text{ terms }$ in the A.P.

$\displaystyle \\$

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

Answer:

$\displaystyle \text{Given } S_7 = 10$

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

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

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

$\displaystyle \text{Also } S_{14}- S_7 = 17$

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

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

$\displaystyle \Rightarrow 14a + 140 d = 34$

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

Solving i) and ii)

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

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

$\displaystyle \Rightarrow 49 d = 7$

$\displaystyle \Rightarrow d = 7$

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

$\displaystyle \text{Therefore } \text{The series is } 1, \frac{8}{7} , \frac{9}{7} , \frac{10}{7} , \ldots$

$\displaystyle \\$

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

Answer:

$\displaystyle \text{Given } a_3 = 7$

$\displaystyle \text{We know } a_n = n + (n-1) d$

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

$\displaystyle \Rightarrow a + 2d = 7$ … … … … … i)

$\displaystyle \text{Also } a_7 - 3a_3 = 2$

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

$\displaystyle \Rightarrow a_7 = 23$

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

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

Solving i) and ii) we get

$\displaystyle 7 - 2d = 23-6d$

$\displaystyle \Rightarrow 4d = 16$

$\displaystyle \Rightarrow d = 4$

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

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

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

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

$\displaystyle \\$

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

Answer:

$\displaystyle \text{Given } a =2 , \ \ \ \ l = 50, \ \ \ \ \ S_n = 442$

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

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

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

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

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

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

$\displaystyle \Rightarrow n = 17$

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

Therefore common difference is $\displaystyle 3 .$

$\displaystyle \\$

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

Answer:

Let the total number of terms be $\displaystyle 2n$

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

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

Subtracting i) from ii) we get

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

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

$\displaystyle \Rightarrow nd = 6$

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

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

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

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

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

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

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

Therefore there are $\displaystyle 8 \text{ terms }$ in the series.

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

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

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

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

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

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

$\displaystyle \Rightarrow 2a = 15 - 12$

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

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

$\displaystyle \\$

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

Answer:

$\displaystyle \text{Given } S_n = n^2p$

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

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

$\displaystyle \text{Also } S_m = m^2 p$

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

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

Solving i) and ii) we get

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

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

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

$\displaystyle \Rightarrow 2p = d$ … … … … … iii)

Substituting back in i) we get

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

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

$\displaystyle \Rightarrow a = p$

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

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

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

$\displaystyle \Rightarrow S_p = p^3$

$\displaystyle \\$

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

Answer:

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

$\displaystyle \text{We know } a_n = a + ( n-1) d$

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

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

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

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

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

Solving i) and ii) we get

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

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

$\displaystyle \Rightarrow -19d = 38$

$\displaystyle \Rightarrow d = - 2$

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

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

$\displaystyle \\$

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

Answer:

$\displaystyle \text{Given } a_{5} = 30 \text{ and } a_{12} = 65$

$\displaystyle \text{We know } a_n = a + ( n-1) d$

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

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

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

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

Solving i) and ii) we get

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

$\displaystyle \Rightarrow 7d = 35$

$\displaystyle \Rightarrow d = 5$

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

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

$\displaystyle \\$

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

Answer:

$\displaystyle \text{Given } a_k = 5k + 1$

$\displaystyle \text{For } k = 1, a_1 = 5(1) + 1 = 6$

$\displaystyle \text{For } k = 2, a_2 = 5(2) + 1 = 11$

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

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

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

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

$\displaystyle \\$

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

Answer:

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

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

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

$\displaystyle \text{We know } a_n = a + ( n-1) d$

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

$\displaystyle \Rightarrow 4n = 88$

$\displaystyle \Rightarrow n = 22$

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

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

$\displaystyle \\$

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

Answer:

$\displaystyle \text{Given series } 25,22,19, \ldots$

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

Question 27: Find the sum of odd integers from $\displaystyle 1 \text{ to } 2001 .$

Answer:

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

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

$\displaystyle \text{We know } a_n = a + ( n-1) d$

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

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

$\displaystyle \Rightarrow n = 1001$

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

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

$\displaystyle \\$

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

Answer:

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

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

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

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

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

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

$\displaystyle -100 = n^2 - 25 n$

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

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

$\displaystyle n = 20$ or $\displaystyle n = 5$

$\displaystyle \\$

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

Answer:

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

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

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

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

Substituting in i)

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

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

$\displaystyle 8d + 8 = 7d + 2$

$\displaystyle d = - 6$

$\displaystyle \text{We know } a_n = a + ( n-1) d$

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

$\displaystyle \\$

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

Answer:

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

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

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

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

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

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

$\displaystyle \\$

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:

$\displaystyle \text{Given } S_n = 3n^2$

$\displaystyle \text{For } n = 1, S_1 = 3(1)^2 = 3$

$\displaystyle \text{For } n = 2 , S_2 = 3(2)^2 = 12$

$\displaystyle \text{For } n = 3, S_3 = (3)^2 = 27$

$\displaystyle \therefore a_1= 3$

$\displaystyle a_2 = 12 - 3 = 9$

$\displaystyle a_3 = 27 - 12 = 15$

$\displaystyle \therefore d = 9 - 3 = 6$

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

$\displaystyle \\$

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

Answer:

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

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

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

$\displaystyle a_1 = P$

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

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

$\displaystyle \\$

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

Answer:

Let for first A.P. $\displaystyle a = a_1 \text{ and } d= d_1$

Let for second A.P. $\displaystyle a = a_2 \text{ and } d= d_2$

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

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

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

$\displaystyle \text{If you put } n = 35 \text{ we get }$

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

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

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

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

$\displaystyle \\$

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