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 ]

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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]

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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)

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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

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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

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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

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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

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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

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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

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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

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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

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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]

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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

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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

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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

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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.

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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

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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

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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 .

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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

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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

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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

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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

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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)

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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

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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

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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

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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

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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

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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.

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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

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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

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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.}}

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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.}}