RD Sharma Class 10 Solutions Chapter 9 Arithmetic Progressions Exercise 9.6 (Updated for 2021-22)

RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6: This exercise is about problems with an A.P.’s sum of terms. RD Sharma Class 10 Solutions is the best material for students to refer to and study effectively for their exams because it covers a wide range of real problems. The following are the RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6 PDF exercise solutions.

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RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6

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RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6 Q2.

Find the sum to n terms of the A.P. 5, 2, –1, – 4, –7, …

Solution:

Given AP is 5, 2, -1, -4, -7, …..

Here, a = 5, d = 2 – 5 = -3

We know that,

S_n = n/2{2a + (n – 1)d}

= n/2{2.5 + (n – 1) – 3}

= n/2{10 – 3(n – 1)}

= n/2{13 – 3n)

∴ S_n = n/2(13 – 3n)

RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6 Q3.

Find the sum of n terms of an A.P. whose the terms is given by a_n = 5 – 6n.

Solution:

Given nth term of the A.P as a_n = 5 – 6n

Put n = 1, we get

a₁ = 5 – 6.1 = -1

So, first term (a) = -1

Last term (a_n) = 5 – 6n = 1

Then, S_n = n/2(-1 + 5 – 6n)

= n/2(4 – 6n) = n(2 – 3n)

RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6 Q4.

Find the sum of last ten terms of the A.P. : 8, 10, 12, 14, .. , 126

Solution:

Given A.P. 8, 10, 12, 14, .. , 126

Here, a = 8 , d = 10 – 8 = 2

We know that, a_n = a + (n – 1)d

So, to find the number of terms

126 = 8 + (n – 1)2

126 = 8 + 2n – 2

2n = 120

n = 60

Next, let’s find the 51^st term

a₅₁ = 8 + 50(2) = 108

So, the sum of last ten terms is the sum of a₅₁ + a₅₂ + a₅₃ + ……. + a₆₀

Here, n = 10, a = 108 and l = 126

S = 10/2 [108 + 126]

= 5(234)

= 1170

Hence, the sum of the last ten terms of the A.P is 1170.

RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6 Q5.

Find the sum of the first 15 terms of each of the following sequences having n^th term as:

(i) a_n = 3 + 4n 

(ii) b_n = 5 + 2n 

(iii) x_n = 6 – n 

(iv) y_n = 9 – 5n

Solution:

(i) Given an A.P. whose n^th term is given by a_n = 3 + 4n

To find the sum of the n terms of the given A.P., using the formula,

S_n = n(a + l)/ 2

Where, a = the first term l = the last term.

Putting n = 1 in the given a_n, we get

a = 3 + 4(1) = 3 + 4 = 7

For the last term (l), here n = 15

a₁₅ = 3 + 4(15) = 63

So, S_n = 15(7 + 63)/2

= 15 x 35

= 525

Therefore, the sum of the 15 terms of the given A.P. is S₁₅ = 525

(ii) Given an A.P. whose n^th term is given by b_n = 5 + 2n

To find the sum of the n terms of the given A.P., using the formula,

S_n = n(a + l)/ 2

Where, a = the first term l = the last term.

Putting n = 1 in the given b_n, we get

a = 5 + 2(1) = 5 + 2 = 7

For the last term (l), here n = 15

a₁₅ = 5 + 2(15) = 35

So, S_n = 15(7 + 35)/2

= 15 x 21

= 315

Therefore, the sum of the 15 terms of the given A.P. is S₁₅ = 315

(iii) Given an A.P. whose n^th term is given by x_n = 6 – n

To find the sum of the n terms of the given A.P., using the formula

S_n = n(a + l)/ 2

Where, a = the first term l = the last term.

Putting n = 1 in the given x_n, we get

a = 6 – 1 = 5

For the last term (l), here n = 15

a₁₅ = 6 – 15 = -9

So, S_n = 15(5 – 9)/2

= 15 x (-2)

= -30

Therefore, the sum of the 15 terms of the given A.P. is S₁₅ = -30

(iv) Given an A.P. whose n^th term is given by y_n = 9 – 5n

To find the sum of the n terms of the given A.P., using the formula,

S_n = n(a + l)/ 2

Where, a = the first term l = the last term.

Putting n = 1 in the given y_n, we get

a = 9 – 5(1) = 9 – 5 = 4

For the last term (l), here n = 15

a₁₅ = 9 – 5(15) = -66

So, S_n = 15(4 – 66)/2

= 15 x (-31)

= -465

Therefore, the sum of the 15 terms of the given A.P. is S₁₅ = -465

RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6 Q6.

Find the sum of the first 20 terms the sequence whose n^th term is a_n = An + B.

Solution:

Given an A.P. whose nth term is given by, a_n = An + B

We need to find the sum of the first 20 terms.

To find the sum of the n terms of the given A.P., we use the formula,

S_n = n(a + l)/ 2

Where, a = the first term l = the last term,

Putting n = 1 in the given a_n, we get

a = A(1) + B = A + B

For the last term (l), here n = 20

A₂₀ = A(20) + B = 20A + B

S₂₀ = 20/2((A + B) + 20A + B)

= 10[21A + 2B]

= 210A + 20B

Therefore, the sum of the first 20 terms of the given A.P. is 210 A + 20B

RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6 Q7.

Find the sum of the first 25 terms of an A.P whose n^th term is given by a_n = 2 – 3n.

Solution:

Given an A.P. whose n^th term is given by a_n = 2 – 3n

To find the sum of the n terms of the given A.P., we use the formula,

S_n = n(a + l)/ 2

Where, a = the first term l = the last term.

Putting n = 1 in the given a_n, we get

a = 2 – 3(1) = -1

For the last term (l), here n = 25

a₂₅ = 2 – 3(25) = -73

So, S_n = 25(-1 – 73)/2

= 25 x (-37)

= -925

Therefore, the sum of the 25 terms of the given A.P. is S₂₅ = -925

RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6 Q8.

Find the sum of the first 25 terms of an A.P whose n^th term is given by a_n = 7 – 3n. 

Solution:

Given an A.P. whose n^th term is given by a_n = 7 – 3n

To find the sum of the n terms of the given A.P., we use the formula,

S_n = n(a + l)/ 2

Where, a = the first term l = the last term.

Putting n = 1 in the given a_n, we get

a = 7 – 3(1) = 7 – 3 = 4

For the last term (l), here n = 25

a₁₅ = 7 – 3(25) = -68

So, S_n = 25(4 – 68)/2

= 25 x (-32)

= -800

Therefore, the sum of the 15 terms of the given A.P. is S₂₅ = -800

RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6 Q9.

If the sum of a certain number of terms starting from the first term of an A.P. is 25, 22, 19, . . ., is 116. Find the last term.

Solution:

Given the sum of the certain number of terms of an A.P. = 116

We know that, S_n = n/2[2a + (n − 1)d]

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms So for the given A.P.(25, 22, 19,…)

Here we have, the first term (a) = 25

The sum of n terms S_n = 116

Common difference of the A.P. (d) = a₂ – a₁ = 22 – 25 = -3

Now, substituting values in S_n

⟹ 116 = n/2[2(25) + (n − 1)(−3)]

⟹ (n/2)[50 + (−3n + 3)] = 116

⟹ (n/2)[53 − 3n] = 116

⟹ 53n – 3n² = 116 x 2

Thus, we get the following quadratic equation,

3n² – 53n + 232 = 0

By factorization method of solving, we have

⟹ 3n² – 24n – 29n + 232 = 0

⟹ 3n( n – 8 ) – 29 ( n – 8 ) = 0

⟹ (3n – 29)( n – 8 ) = 0

So, 3n – 29 = 0

⟹ n = 29/3

Also, n – 8 = 0

⟹ n = 8

Since n cannot be a fraction, so the number of terms is taken as 8.

So, the term is:

a₈ = a₁ + 7d = 25 + 7(-3) = 25 – 21 = 4

Hence, the last term of the given A.P. such that the sum of the terms is 116 is 4.

RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6 Q10.

(i) How many terms of the sequence 18, 16, 14…. should be taken so that their sum is zero. 

(ii) How many terms are there in the A.P. whose first and fifth terms are -14 and 2 respectively and the sum of the terms is 40? 

(iii) How many terms of the A.P. 9, 17, 25, . . . must be taken so that their sum is 636? 

(iv) How many terms of the A.P. 63, 60, 57, . . . must be taken so that their sum is 693? 

(v) How many terms of the A.P. is 27, 24, 21. . . should be taken that their sum is zero? 

Solution:

(i) Given AP. is 18, 16, 14, …

We know that,

S_n = n/2[2a + (n − 1)d]

Here,

The first term (a) = 18

The sum of n terms (S_n) = 0 (given)

Common difference of the A.P.

(d) = a₂ – a₁ = 16 – 18 = – 2

So, on substituting the values in S_n

⟹ 0 = n/2[2(18) + (n − 1)(−2)]

⟹ 0 = n/2[36 + (−2n + 2)]

⟹ 0 = n/2[38 − 2n] Further, n/2

⟹ n = 0 Or, 38 – 2n = 0

⟹ 2n = 38

⟹ n = 19

Since the number of terms cannot be zer0, hence the number of terms (n) should be 19.

(ii) Given, the first term (a) = -14, Filth term (a₅) = 2, Sum of terms (S_n) = 40 of the A.P.

If the common difference is taken as d.

Then, a₅ = a+ 4d

⟹ 2 = -14 + 4d

⟹ 2 + 14 = 4d

⟹ 4d = 16

⟹ d = 4

Next, we know that S_n = n/2[2a + (n − 1)d]

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

Now, on substituting the values in S_n

⟹ 40 = n/2[2(−14) + (n − 1)(4)]

⟹ 40 = n/2[−28 + (4n − 4)]

⟹ 40 = n/2[−32 + 4n]

⟹ 40(2) = – 32n + 4n²

So, we get the following quadratic equation,

4n² – 32n – 80 = 0

⟹ n² – 8n – 20 = 0

On solving by factorization method, we get

n² – 10n + 2n – 20 = 0

⟹ n(n – 10) + 2(n – 10 ) = 0

⟹ (n + 2)(n – 10) = 0

Either, n + 2 = 0

⟹ n = -2

Or, n – 10 = 0

⟹ n = 10

Since the number of terms cannot be negative.

Therefore, the number of terms (n) is 10.

(iii) Given AP is 9, 17, 25,…

We know that,

S_n = n/2[2a + (n − 1)d]

Here we have,

The first term (a) = 9 and the sum of n terms (S_n) = 636

Common difference of the A.P. (d) = a₂ – a₁ = 17 – 9 = 8

Substituting the values in S_n, we get

⟹ 636 = n/2[2(9) + (n − 1)(8)]

⟹ 636 = n/2[18 + (8n − 8)]

⟹ 636(2) = (n)[10 + 8n]

⟹ 1271 = 10n + 8n²

Now, we get the following quadratic equation,

⟹ 8n² + 10n – 1272 = 0

⟹ 4n²+ 5n – 636 = 0

On solving by factorisation method, we have

⟹ 4n² – 48n + 53n – 636 = 0

⟹ 4n(n – 12) + 53(n – 12) = 0

⟹ (4n + 53)(n – 12) = 0

Either 4n + 53 = 0 ⟹ n = -53/4

Or, n – 12 = 0 ⟹ n = 12

Since, the number of terms cannot be a fraction.

Therefore, the number of terms (n) is 12.

(iv) Given A.P. is 63, 60, 57,…

We know that,

S_n = n/2[2a + (n − 1)d]

Here we have,

the first term (a) = 63

The sum of n terms (S_n) = 693

Common difference of the A.P. (d) = a₂ – a₁ = 60 – 63 = –3

On substituting the values in S_n we get

⟹ 693 = n/2[2(63) + (n − 1)(−3)]

⟹ 693 = n/2[126+(−3n + 3)]

⟹ 693 = n/2[129 − 3n]

⟹ 693(2) = 129n – 3n²

Now, we get the following quadratic equation.

⟹ 3n² – 129n + 1386 = 0

⟹ n² – 43n + 462

Solving by factorisation method, we have

⟹ n² – 22n – 21n + 462 = 0

⟹ n(n – 22) -21(n – 22) = 0

⟹ (n – 22) (n – 21) = 0

Either, n – 22 = 0 ⟹ n = 22

Or, n – 21 = 0 ⟹ n = 21

Now, the 22^nd term will be a₂₂ = a₁ + 21d = 63 + 21( -3 ) = 63 – 63 = 0

So, the sum of 22 as well as 21 terms is 693.

Therefore, the number of terms (n) is 21 or 22.

(v) Given A.P. is 27, 24, 21. . .

We know that,

S_n = n/2[2a + (n − 1)d]

Here we have, the first term (a) = 27

The sum of n terms (S_n) = 0

Common difference of the A.P. (d) = a₂ – a₁ = 24 – 27 = -3

On substituting the values in S_n, we get

⟹ 0 = n/2[2(27) + (n − 1)( − 3)]

⟹ 0 = (n)[54 + (n – 1)(-3)]

⟹ 0 = (n)[54 – 3n + 3]

⟹ 0 = n [57 – 3n] Further we have, n = 0 Or, 57 – 3n = 0

⟹ 3n = 57

⟹ n = 19

The number of terms cannot be zero,

Hence, the numbers of terms (n) is 19.

RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6 Q11.

Find the sum of the first

(i) 11 terms of the A.P. : 2, 6, 10, 14, . . . 

(ii) 13 terms of the A.P. : -6, 0, 6, 12, . . . 

(iii) 51 terms of the A.P.: whose second term is 2 and the fourth term is 8.

Solution:

We know that the sum of terms for different arithmetic progressions is given by

S_n = n/2[2a + (n − 1)d]

Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms

(i) Given A.P 2, 6, 10, 14,… to 11 terms.

Common difference (d) = a₂ – a₁ = 10 – 6 = 4

Number of terms (n) = 11

First term for the given A.P. (a) = 2

So,

S₁₁ = 11/2[2(2) + (11 − 1)4]

= 11/2[2(2) + (10)4]

= 11/2[4 + 40]

= 11 × 22

= 242

Hence, the sum of first 11 terms for the given A.P. is 242

(ii) Given A.P. – 6, 0, 6, 12, … to 13 terms.

Common difference (d) = a₂ – a₁ = 6 – 0 = 6

Number of terms (n) = 13

First term (a) = -6

So,

S₁₃ = 13/2[2(− 6) + (13 –1)6]

= 13/2[(−12) + (12)6]

= 13/2[60] = 390

Hence, the sum of first 13 terms for the given A.P. is 390

(iii) 51 terms of an AP whose a₂ = 2 and a₄ = 8

We know that, a₂ = a + d

2 = a + d …(2)

Also, a₄ = a + 3d

8 = a + 3d … (2)

Subtracting (1) from (2), we have

2d = 6

d = 3

Substituting d = 3 in (1), we get

2 = a + 3

⟹ a = -1

Given that the number of terms (n) = 51

First term (a) = -1

So,

S_n = 51/2[2(−1) + (51 − 1)(3)]

= 51/2[−2 + 150]

= 51/2[148]

= 3774

Hence, the sum of the first 51 terms for the A.P. is 3774.

RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6 Q12.

Find the sum of

(i) the first 15 multiples of 8

(ii) the first 40 positive integers divisible by (a) 3 (b) 5 (c) 6.

(iii) all 3 – digit natural numbers which are divisible by 13.

(iv) all 3 – digit natural numbers which are multiples of 11.

Solution:

We know that the sum of terms for an A.P is given by

S_n = n/2[2a + (n − 1)d]

Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms

(i) Given, first 15 multiples of 8.

These multiples form an A.P: 8, 16, 24, ……, 120

Here, a = 8 , d = 61 – 8 = 8 and the number of terms(n) = 15

Now, finding the sum of 15 terms, we have

\

Hence, the sum of the first 15 multiples of 8 is 960

(ii)(a) First 40 positive integers divisible by 3.

Hence, the first multiple is 3 and the 40^th multiple is 120.

And, these terms will form an A.P. with the common difference of 3.

Here, First term (a) = 3

Number of terms (n) = 40

Common difference (d) = 3

So, the sum of 40 terms

S₄₀ = 40/2[2(3) + (40 − 1)3]

= 20[6 + (39)3]

= 20(6 + 117)

= 20(123) = 2460

Thus, the sum of the first 40 multiples of 3 is 2460.

(b) First 40 positive integers divisible by 5

Hence, the first multiple is 5 and the 40^th multiple is 200.

And, these terms will form an A.P. with the common difference of 5.

Here, First term (a) = 5

Number of terms (n) = 40

Common difference (d) = 5

So, the sum of 40 terms

S₄₀ = 40/2[2(5) + (40 − 1)5]

= 20[10 + (39)5]

= 20 (10 + 195)

= 20 (205) = 4100

Hence, the sum of the first 40 multiples of 5 is 4100.

(c) First 40 positive integers divisible by 6

Hence, the first multiple is 6 and the 40^th multiple is 240.

And, these terms will form an A.P. with the common difference of 6.

Here, First term (a) = 6

Number of terms (n) = 40

Common difference (d) = 6

So, the sum of 40 terms

S₄₀ = 40/2[2(6) + (40 − 1)6]

= 20[12 + (39)6]

=20(12 + 234)

= 20(246) = 4920

Hence, the sum of the first 40 multiples of 6 is 4920.

(iii) All 3 digit natural numbers are divisible by 13.

So, we know that the first 3 digit multiple of 13 is 104 and the last 3 digit multiple of 13 is 988.

And, these terms form an A.P. with the common difference of 13.

Here, first term (a) = 104 and the last term (l) = 988

Common difference (d) = 13

Finding the number of terms in the A.P. by, a_n = a + (n − 1)d

We have,

988 = 104 + (n – 1)13

⟹ 988 = 104 + 13n -13

⟹ 988 = 91 + 13n

⟹ 13n = 897

⟹ n = 69

Now, using the formula for the sum of n terms, we get

S₆₉ = 69/2[2(104) + (69 − 1)13]

= 69/2[208 + 884]

= 69/2[1092]

= 69(546)

= 37674

Hence, the sum of all 3 digit multiples of 13 is 37674.

(iv) All 3 digit natural numbers which are multiples of 11.

So, we know that the first 3 digit multiple of 11 is 110 and the last 3 digit multiple of 13 is 990.

And, these terms form an A.P. with the common difference of 11.

Here, first term (a) = 110 and the last term (l) = 990

Common difference (d) = 11

Finding the number of terms in the A.P. by, a_n = a + (n − 1)d

We get,

990 = 110 + (n – 1)11

⟹ 990 = 110 + 11n -11

⟹ 990 = 99 + 11n

⟹ 11n = 891

⟹ n = 81

Now, using the formula for the sum of n terms, we get

S₈₁ = 81/2[2(110) + (81 − 1)11]

= 81/2[220 + 880]

= 81/2[1100]

= 81(550)

= 44550

Hence, the sum of all 3 digit multiples of 11 is 44550.

RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6 Q13.

Find the sum:

(i) 2 + 4 + 6 + . . . + 200 

(ii) 3 + 11 + 19 + . . . + 803 

(iii) (-5) + (-8) + (-11) + . . . + (- 230) 

(iv) 1 + 3 + 5 + 7 + . . . + 199 

(vi) 34 + 32 + 30 + . . . + 10 

(vii) 25 + 28 + 31 + . . . + 100 

Solution:

We know that the sum of terms for an A.P is given by

S_n = n/2[2a + (n − 1)d]

Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms

Or S_n = n/2[a + l]

Where; a = first term for the given A.P. ;l = last term for the given A.P

(i) Given series. 2 + 4 + 6 + . . . + 200 which is an A.P

Where, a = 2 ,d = 4 – 2 = 2 and last term (a_n = l) = 200

We know that, a_n = a + (n – 1)d

So,

200 = 2 + (n – 1)2

200 = 2 + 2n – 2

n = 200/2 = 100

Now, for the sum of these 100 terms

S₁₀₀ = 100/2 [2 + 200]

= 50(202)

= 10100

Hence, the sum of terms of the given series is 10100.

(ii) Given series. 3 + 11 + 19 + . . . + 803 which is an A.P

Where, a = 3 ,d = 11 – 3 = 8 and last term (a_n = l) = 803

We know that, a_n = a + (n – 1)d

So,

803 = 3 + (n – 1)8

803 = 3 + 8n – 8

n = 808/8 = 101

Now, for the sum of these 101 terms

S₁₀₁ = 101/2 [3 + 803]

= 101(806)/2

= 101 x 403

= 40703

Hence, the sum of terms of the given series is 40703.

(iii) Given series (-5) + (-8) + (-11) + . . . + (- 230)which is an A.P

Where, a = -5 ,d = -8 – (-5) = -3 and last term (a_n = l) = -230

We know that, a_n = a + (n – 1)d

So,

-230 = -5 + (n – 1)(-3)

-230 = -5 – 3n + 3

3n = -2 + 230

n = 228/3 = 76

Now, for the sum of these 76 terms

S₇₆ = 76/2 [-5 + (-230)]

= 38 x (-235)

= -8930

Hence, the sum of terms of the given series is -8930.

(iv) Given series. 1 + 3 + 5 + 7 + . . . + 199which is an A.P

Where, a = 1 ,d = 3 – 1 = 2 and last term (a_n = l) = 199

We know that, a_n = a + (n – 1)d

So,

199 = 1 + (n – 1)2

199 = 1 + 2n – 2

n = 200/2 = 100

Now, for the sum of these 100 terms

S₁₀₀ = 100/2 [1 + 199]

= 50(200)

= 10000

Hence, the sum of terms of the given series is 10000.

(v) Given series which is an A.P

Where, a = 7, d = 10 ½ – 7 = (21 – 14)/2 = 7/2 and last term (a_n = l) = 84

We know that, a_n = a + (n – 1)d

So,

84 = 7 + (n – 1)(7/2)

168 = 14 + 7n – 7

n = (168 – 7)/7 = 161/7 = 23

Now, for the sum of these 23 terms

S₂₃ = 23/2 [7 + 84]

= 23(91)/2

= 2093/2

Hence, the sum of terms of the given series is 2093/2.

(vi) Given series, 34 + 32 + 30 + . . . + 10which is an A.P

Where, a = 34 ,d = 32 – 34 = -2 and last term (a_n = l) = 10

We know that, a_n = a + (n – 1)d

So,

10 = 34 + (n – 1)(-2)

10 = 34 – 2n + 2

n = (36 – 10)/2 = 13

Now, for the sum of these 13 terms

S₁₃ = 13/2 [34 + 10]

= 13(44)/2

= 13 x 22

= 286

Hence, the sum of terms of the given series is 286.

(vii) Given series, 25 + 28 + 31 + . . . + 100which is an A.P

Where, a = 25 ,d = 28 – 25 = 3 and last term (a_n = l) = 100

We know that, a_n = a + (n – 1)d

So,

100 = 25 + (n – 1)(3)

100 = 25 + 3n – 3

n = (100 – 22)/3 = 26

Now, for the sum of these 26 terms

S₁₀₀ = 26/2 [25 + 100]

= 13(125)

= 1625

Hence, the sum of terms of the given series is 1625.

RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6 Q14.

The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

Solution:

Given, the first term of the A.P (a) = 17

The last term of the A.P (l) = 350

The common difference (d) of the A.P. = 9

Let the number of terms be n. And, we know that; l = a + (n – 1)d

So, 350 = 17 + (n- 1) 9

⟹ 350 = 17 + 9n – 9

⟹ 350 = 8 + 9n

⟹ 350 – 8 = 9n

Thus we get, n = 38

Now, finding the sum of terms

S_n = n/2[a + l]

= 38/2(17 + 350)

= 19 × 367

= 6973

Hence, the number of terms is of the A.P is 38 and their sum is 6973.

RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6 Q15.

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 the first 20 terms.

Solution:

Let’s consider the first term as a and the common difference as d.

Given,

a₃ = 7 …. (1) and,

a₇ = 3a₃ + 2 …. (2)

So, using (1) in (2), we get,

a₇ = 3(7) + 2 = 21 + 2 = 23 …. (3)

Also, we know that

a_n = a +(n – 1)d

So, the 3th term (for n = 3),

a₃ = a + (3 – 1)d

⟹ 7 = a + 2d (Using 1)

⟹ a = 7 – 2d …. (4)

Similarly, for the 7th term (n = 7),

a₇ = a + (7 – 1) d 24 = a + 6d = 23 (Using 3)

a = 23 – 6d …. (5)

Subtracting (4) from (5), we get,

a – a = (23 – 6d) – (7 – 2d)

⟹ 0 = 23 – 6d – 7 + 2d

⟹ 0 = 16 – 4d

⟹ 4d = 16

⟹ d = 4

Now, to find a, we substitute the value of d in (4), a =7 – 2(4)

⟹ a = 7 – 8

a = -1

Hence, for the A.P. a = -1 and d = 4

For finding the sum, we know that

S_n = n/2[2a + (n − 1)d] and n = 20 (given)

S₂₀ = 20/2[2(−1) + (20 − 1)(4)]

= (10)[-2 + (19)(4)]

= (10)[-2 + 76]

= (10)[74]

= 740

Hence, the sum of the first 20 terms for the given A.P. is 740

RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6 Q16.

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.

Solution:

Given,

The first term of the A.P (a) = 2

The last term of the A.P (l) = 50

Sum of all the terms S_n = 442

So, let the common difference of the A.P. be taken as d.

The sum of all the terms is given as,

442 = (n/2)(2 + 50)

⟹ 442 = (n/2)(52)

⟹ 26n = 442

⟹ n = 17

Now, the last term is expressed as

50 = 2 + (17 – 1)d

⟹ 50 = 2 + 16d

⟹ 16d = 48

⟹ d = 3

Thus, the common difference of the A.P. is d = 3.

RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6 Q17.

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

Solution:

Let us take the first term as a and the common difference as d.

Given,

a₁₂ = -13 S₄ = 24

Also, we know that a_n = a + (n – 1)d

So, for the 12th term

a₁₂ = a + (12 – 1)d = -13

⟹ a + 11d = -13

a = -13 – 11d …. (1)

And, we that for sum of terms

S_n = n/2[2a + (n − 1)d]

Here, n = 4

S₄ = 4/2[2(a) + (4 − 1)d]

⟹ 24 = (2)[2a + (3)(d)]

⟹ 24 = 4a + 6d

⟹ 4a = 24 – 6d

Subtracting (1) from (2), we have

Further simplifying for d, we get,

⟹ -19 × 2 = 19d

⟹ d = – 2

On substituting the value of d in (1), we find a

a = -13 – 11(-2)

a = -13 + 22

a = 9

Next, the sum of 10 term is given by

S₁₀ = 10/2[2(9) + (10 − 1)(−2)]

= (5)[19 + (9)(-2)]

= (5)(18 – 18) = 0

Thus, the sum of the first 10 terms for the given A.P. is S₁₀ = 0.

18.

19. In an A.P., if the first term is 22, the common difference is – 4, and the sum to n terms is 64, find n.

Solution:

Given that,

a = 22, d = – 4 and S_n = 64

Let us consider the number of terms as n.

For sum of terms in an A.P, we know that

S_n = n/2[2a + (n − 1)d]

Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms

So,

⟹ S_n = n/2[2(22) + (n − 1)(−4)]

⟹ 64 = n/2[2(22) + (n − 1)(−4)]

⟹ 64(2) = n(48 – 4n)

⟹ 128 = 48n – 4n²

After rearranging the terms, we have a quadratic equation

4n² – 48n + 128 = 0,

n² – 12n + 32 = 0 [dividing by 4 on both sides]

n² – 12n + 32 = 0

Solving by factorisation method,

n² – 8n – 4n + 32 = 0

n ( n – 8 ) – 4 ( n – 8 ) = 0

(n – 8) (n – 4) = 0

So, we get n – 8 = 0 ⟹ n = 8

Or, n – 4 = 0 ⟹ n = 4

Hence, the number of terms can be either n = 4 or 8.

20. In an A.P., if the 5^th and 12^th terms are 30 and 65 respectively, what is the sum of first 20 terms? 

Solution:

Let’s take the first term as a and the common difference to be d

Given that,

a₅ = 30 and a₁₂ = 65

And, we know that a_n = a + (n – 1)d

So,

a₅ = a + (5 – 1)d

30 = a + 4d

a = 30 – 4d …. (i)

Similarly, a₁₂ = a + (12 – 1) d

65 = a + 11d

a = 65 – 11d …. (ii)

Subtracting (i) from (ii), we have

a – a = (65 – 11d) – (30 – 4d)

0 = 65 – 11d – 30 + 4d

0 = 35 – 7d

7d = 35

d = 5

Putting d in (i), we get

a = 30 – 4(5)

a = 30 – 20

a = 10

Thus for the A.P; d = 5 and a = 10

Next, to find the sum of first 20 terms of this A.P., we use the following formula for the sum of n terms of an A.P.,

S_n = n/2[2a + (n − 1)d]

Where;

a = first term of the given A.P.

d = common difference of the given A.P.

n = number of terms

Here n = 20, so we have

S₂₀ = 20/2[2(10) + (20 − 1)(5)]

= (10)[20 + (19)(5)]

= (10)[20 + 95]

= (10)[115]

= 1150

Hence, the sum of the first 20 terms for the given A.P. is 1150

21. Find the sum of the first 51 terms of an A.P. whose second and third terms are 14 and 18 respectively.

Solution:

Let’s take the first term as a and the common difference as d.

Given that,

a₂ = 14 and a₃ = 18

And, we know that a_n = a + (n – 1)d

So,

a₂ = a + (2 – 1)d

⟹ 14 = a + d

⟹ a = 14 – d …. (i)

Similarly,

a₃ = a + (3 – 1)d

⟹ 18 = a + 2d

⟹ a = 18 – 2d …. (ii)

Subtracting (i) from (ii), we have

a – a = (18 – 2d) – (14 – d)

0 = 18 – 2d – 14 + d

0 = 4 – d

d = 4

Putting d in (i), to find a

a = 14 – 4

a = 10

Thus, for the A.P. d = 4 and a = 10

Now, to find sum of terms

S_n = n/2(2a + (n − 1)d)

Where,

a = the first term of the A.P.

d = common difference of the A.P.

n = number of terms So, using the formula for

n = 51,

⟹ S₅₁ = 51/2[2(10) + (51 – 1)(4)]

= 51/2[20 + (40)4]

= 51/2[220]

= 51(110)

= 5610

Hence, the sum of the first 51 terms of the given A.P. is 5610

22. If the sum of 7 terms of an A.P. is 49 and that of 17 terms is 289, find the sum of n terms.

Solution:

Given,

The Sum of 7 terms of an A.P. is 49

⟹ S₇ = 49

And, the sum of 17 terms of an A.P. is 289

⟹ S₁₇ = 289

Let the first term of the A.P be a and common difference as d.

And, we know that the sum of n terms of an A.P is

S_n = n/2[2a + (n − 1)d]

So,

S₇ = 49 = 7/2[2a + (7 – 1)d]

= 7/2 [2a + 6d]

= 7[a + 3d]

⟹ 7a + 21d = 49

a + 3d = 7 ….. (i)

Similarly,

S₁₇ = 17/2[2a + (17 – 1)d]

= 17/2 [2a + 16d]

= 17[a + 8d]

⟹ 17[a + 8d] = 289

a + 8d = 17 ….. (ii)

Now, subtracting (i) from (ii), we have

a + 8d – (a + 3d) = 17 – 7

5d = 10

d = 2

Putting d in (i), we find a

a + 3(2) = 7

a = 7 – 6 = 1

So, for the A.P: a = 1 and d = 2

For the sum of n terms is given by,

S_n = n/2[2(1) + (n − 1)(2)]

= n/2[2 + 2n – 2]

= n/2[2n]

= n²

Therefore, the sum of n terms of the A.P is given by n².

23. The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

Solution:

The Sum of the first n terms of an A.P is given by S_n = n/2(2a + (n − 1)d)

Given,

First term (a) = 5, last term (a_n) = 45 and sum of n terms (S_n) = 400

Now, we know that

a_n = a + (n – 1)d

⟹ 45 = 5 + (n – 1)d

⟹ 40 = nd – d

⟹ nd – d = 40 …. (1)

Also,

S_n = n/2(2(a) + (n − 1)d)

400 = n/2(2(5) + (n − 1)d)

800 = n (10 + nd – d)

800 = n (10 + 40) [using (1)]

⟹ n = 16

Putting n in (1), we find d

nd – d = 40

16d – d = 40

15d = 40

d = 8/3

Therefore, the common difference of the given A.P. is 8/3.

24. In an A.P. the first term is 8, the n^th term is 33 and the sum of the first n term is 123. Find n and the d, the common difference. 

Solution:

Given,

The first term of the A.P (a) = 8

The nth term of the A.P (l) = 33

And, the sum of all the terms S_n = 123

Let the common difference of the A.P. be d.

So, find the number of terms by

123 = (n/2)(8 + 33)

123 = (n/2)(41)

n = (123 x 2)/ 41

n = 246/41

n = 6

Next, to find the common difference of the A.P. we know that

l = a + (n – 1)d

33 = 8 + (6 – 1)d

33 = 8 + 5d

5d = 25

d = 5

Thus, the number of terms is n = 6 and the common difference of the A.P. is d = 5.

25. In an A.P. the first term is 22, the n^th term is -11 and the sum of the first n term is 66. Find n and the d, the common difference. 

Solution:

Given,

The first term of the A.P (a) = 22

The nth term of the A.P (l) = -11

And, sum of all the terms S_n = 66

Let the common difference of the A.P. be d.

So, finding the number of terms by

66 = (n/2)[22 + (−11)]

66 = (n/2)[22 − 11]

(66)(2) = n(11)

6 × 2 = n

n = 12

Now, for finding d

We know that, l = a + (n – 1)d

– 11 = 22 + (12 – 1)d

-11 = 22 + 11d

11d = – 33

d = – 3

Hence, the number of terms is n = 12, and the common difference d = -3

26. The first and the last terms of an A.P. are 7 and 49 respectively. If the sum of all its terms is 420, find the common difference.

Solution:

Given,

First term (a) = 7, last term (a_n) = 49 and sum of n terms (S_n) = 420

Now, we know that

a_n = a + (n – 1)d

⟹ 49 = 7 + (n – 1)d

⟹ 43 = nd – d

⟹ nd – d = 42 ….. (1)

Next,

S_n = n/2(2(7) + (n − 1)d)

⟹ 840 = n[14 + nd – d]

⟹ 840 = n[14 + 42] [using (1)]

⟹ 840 = 54n

⟹ n = 15 …. (2)

So, by substituting (2) in (1), we have

nd – d = 42

⟹ 15d – d = 42

⟹ 14d = 42

⟹ d = 3

Therefore, the common difference of the given A.P. is 3.

27. The first and the last terms of an A.P are 5 and 45 respectively. If the sum of all its terms is 400, find its common difference.

Solution:

Given,

First term (a) = 5 and the last term (l) = 45

Also, S_n = 400

We know that,

a_n = a + (n – 1)d

⟹ 45 = 5 + (n – 1)d

⟹ 40 = nd – d

⟹ nd – d = 40 ….. (1)

Next,

S_n = n/2(2(5) + (n − 1)d)

⟹ 400 = n[10 + nd – d]

⟹ 800 = n[10 + 40] [using (1)]

⟹ 800 = 50n

⟹ n = 16 …. (2)

So, by substituting (2) in (1), we have

nd – d = 40

⟹ 16d – d = 40

⟹ 15d = 40

⟹ d = 8/3

Therefore, the common difference of the given A.P. is 8/3.

28. The sum of the first q terms of an A.P. is 162. The ratio of its 6^th term to its 13^th term is 1: 2. Find the first and 15^th terms of the A.P.

Solution:

Let a be the first term and d be a common difference.

And we know that, sum of first n terms is:

S_n = n/2(2a + (n − 1)d)

Also, nth term is given by a_n = a + (n – 1)d

From the question, we have

S_q = 162 and a₆ : a₁₃ = 1 : 2

So,

2a₆ = a₁₃

⟹ 2 [a + (6 – 1d)] = a + (13 – 1)d

⟹ 2a + 10d = a + 12d

⟹ a = 2d …. (1)

And, S₉ = 162

⟹ S₉ = 9/2(2a + (9 − 1)d)

⟹ 162 = 9/2(2a + 8d)

⟹ 162 × 2 = 9[4d + 8d] [from (1)]

⟹ 324 = 9 × 12d

⟹ d = 3

⟹ a = 2(3) [from (1)]

⟹ a = 6

Hence, the first term of the A.P. is 6

For the 15^th term, a₁₅ = a + 14d = 6 + 14 × 3 = 6 + 42

Therefore, a₁₅ = 48

29. If the 10^th term of an A.P. is 21 and the sum of its first 10 terms is 120, find its n^th term.

Solution:

Let’s consider a to be the first term and d be a common difference.

And we know that, sum of first n terms is:

S_n = n/2(2a + (n − 1)d) and n^th term is given by: a_n = a + (n – 1)d

Now, from the question we have

S₁₀ = 120

⟹ 120 = 10/2(2a + (10 − 1)d)

⟹ 120 = 5(2a + 9d)

⟹ 24 = 2a + 9d …. (1)

Also given that, a₁₀ = 21

⟹ 21 = a + (10 – 1)d

⟹ 21 = a + 9d …. (2)

Subtracting (2) from (1), we get

24 – 21 = 2a + 9d – a – 9d

⟹a = 3

Now, on putting a = 3 in equation (2), we get

3 + 9d = 21

9d = 18

d = 2

Thus, we have the first term(a) = 3 and the common difference(d) = 2

Therefore, the n^th term is given by

a_n = a + (n – 1)d = 3 + (n – 1)2

= 3 + 2n -2

= 2n + 1

Hence, the n^th term of the A.P is (a_n) = 2n + 1.

30. The sum of the first 7 terms of an A.P. is 63 and the sum of its next 7 terms is 161. Find the 28^th term of this A.P.

Solution:

Let’s take a to be the first term and d to be a common difference.

And we know that, the sum of first n terms

S_n = n/2(2a + (n − 1)d)

Given that sum of the first 7 terms of an A.P. is 63.

S₇ = 63

And sum of next 7 terms is 161.

So, the sum of first 14 terms = Sum of first 7 terms + sum of next 7 terms

S₁₄ = 63 + 161 = 224

Now, having

S₇ = 7/2(2a + (7 − 1)d)

⟹ 63(2) = 7(2a + 6d)

⟹ 9 × 2 = 2a + 6d

⟹ 2a + 6d = 18 . . . . (1)

And,

S₁₄ = 14/2(2a + (14 − 1)d)

⟹ 224 = 7(2a + 13d)

⟹ 32 = 2a + 13d …. (2)

Now, subtracting (1) from (2), we get

⟹ 13d – 6d = 32 – 18

⟹ 7d = 14

⟹ d = 2

Using d in (1), we have

2a + 6(2) = 18

2a = 18 – 12

a = 3

Thus, from n^th term

⟹ a₂₈ = a + (28 – 1)d

= 3 + 27 (2)

= 3 + 54 = 57

Therefore, the 28^th term is 57.

31. The sum of the first seven terms of an A.P. is 182. If its 4^th and 17^th terms are in ratio 1: 5, find the A.P.

Solution:

Given that,

S₁₇ = 182

And, we know that the sum of first n term is:

S_n = n/2(2a + (n − 1)d)

So,

S₇ = 7/2(2a + (7 − 1)d)

182 × 2 = 7(2a + 6d)

364 = 14a + 42d

26 = a + 3d

a = 26 – 3d … (1)

Also, it’s given that 4^th term and 17^th term are in a ratio of 1: 5. So, we have

⟹ 5(a₄) = 1(a₁₇)

⟹ 5 (a + 3d) = 1 (a + 16d)

⟹ 5a + 15d = a + 16d

⟹ 4a = d …. (2)

Now, substituting (2) in (1), we get

⟹ 4 ( 26 – 3d ) = d

⟹ 104 – 12d = d

⟹ 104 = 13d

⟹ d = 8

Putting d in (2), we get

⟹ 4a = d

⟹ 4a = 8

⟹ a = 2

Therefore, the first term is 2 and the common difference is 8. So, the A.P. is 2, 10, 18, 26, . ..

32. The n^th term of an A.P is given by (-4n + 15). Find the sum of the first 20 terms of this A.P.

Solution:

Given,

The n^th term of the A.P = (-4n + 15)

So, by putting n = 1 and n = 20 we can find the first ans 20^th term of the A.P

a = (-4(1) + 15) = 11

And,

a₂₀ = (-4(20) + 15) = -65

Now, for find the sum of 20 terms of this A.P we have the first and last term.

So, using the formula

S_n = n/2(a + l)

S₂₀ = 20/2(11 + (-65))

= 10(-54)

= -540

Therefore, the sum of the first 20 terms of this A.P. is -540.

33. In an A.P. the sum of the first ten terms is -150 and the sum of its next 10 terms is -550. Find the A.P.

Solution:

Let’s take a to be the first term and d to be a common difference.

And we know that, the sum of first n terms

S_n = n/2(2a + (n − 1)d)

Given that sum of the first 10 terms of an A.P. is -150.

S₁₀ = -150

And the sum of next 10 terms is -550.

So, the sum of first 20 terms = Sum of first 10 terms + sum of next 10 terms

S₂₀ = -150 + -550 = -700

Now, having

S₁₀ = 10/2(2a + (10 − 1)d)

⟹ -150 = 5(2a + 9d)

⟹ -30 = 2a + 9d

⟹ 2a + 9d = -30 . . . . (1)

And,

S₂₀ = 20/2(2a + (20 − 1)d)

⟹ -700 = 10(2a + 19d)

⟹ -70 = 2a + 19d …. (2)

Now, subtracting (1) from (2), we get

⟹ 19d – 9d = -70 – (-30)

⟹ 10d = -40

⟹ d = -4

Using d in (1), we have

2a + 9(-4) = -30

2a = -30 + 36

a = 6/2 = 3

Hence, we have a = 3 and d = -4

So, the A.P is 3, -1, -5, -9, -13,…..

34. Sum of the first 14 terms of an A.P. is 1505 and its first term is 10. Find its 25^th term.

Solution

Given,

First term of the A.P is 1505 and

S₁₄ = 1505

We know that, the sum of first n terms is

S_n = n/2(2a + (n − 1)d)

So,

S₁₄ = 14/2(2(10) + (14 − 1)d) = 1505

7(20 + 13d) = 1505

20 + 13d = 215

13d = 215 – 20

d = 195/13

d =15

Thus, the 25^th term is given by

a₂₅ = 10 + (25 -1)15

= 10 + (24)15

= 10 + 360

= 370

Therefore, the 25^th term of the A.P is 370

35. In an A.P. , the first term is 2, the last term is 29 and the sum of the terms is 155. Find the common difference of the A.P.

Solution:

Given,

The first term of the A.P. (a) = 2

The last term of the A.P. (l) = 29

And, sum of all the terms (S_n) = 155

Let the common difference of the A.P. be d.

So, find the number of terms by sum of terms formula

S_n = n/2 (a + l)

155 = n/2(2 + 29)

155(2) = n(31)

31n = 310

n = 10

Using n for the last term, we have

l = a + (n – 1)d

29 = 2 + (10 – 1)d

29 = 2 + (9)d

29 – 2 = 9d

9d = 27

d = 3

Hence, the common difference of the A.P. is d = 3

36. The first and the last term of an A.P are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

Solution:

Given,

In an A.P first term (a) = 17 and the last term (l) = 350

And, the common difference (d) = 9

We know that,

a_n = a + (n – 1)d

so,

a_n = l = 17 + (n – 1)9 = 350

17 + 9n – 9 = 350

9n = 350 – 8

n = 342/9

n = 38

So, the sum of all the terms of the A.P is given by

S_n = n/2 (a + l)

= 38/2(17 + 350)

= 19(367)

= 6973

Therefore, the sum of terms of the A.P is 6973.

37. Find the number of terms of the A.P. –12, –9, –6, . . . , 21. If 1 is added to each term of this A.P., then find the sum of all terms of the A.P. thus obtained.

Solution:

Given,

First term, a = -12

Common difference, d = a₂ – a₁ = – 9 – (- 12)

d = – 9 + 12 = 3

And, we know that n^th term = a_n = a + (n – 1)d

⟹ 21 = -12 + (n – 1)3

⟹ 21 = -12 + 3n – 3

⟹ 21 = 3n – 15

⟹ 36 = 3n

⟹ n = 12

Thus, the number of terms is 12.

Now, if 1 is added to each of the 12 terms, the sum will increase by 12.

Hence, the sum of all the terms of the A.P. so obtained is

⟹ S₁₂ + 12 = 12/2[a + l] + 12

= 6[-12 + 21] + 12

= 6 × 9 + 12

= 66

Therefore, the sum after adding 1 to each of the terms in the A.P is 66.

We have provided complete details of RD Sharma Class 10 Solutions Chapter 9 Exercise 9.6. If you have any queries related to CBSE Class 10, feel free to ask us in the comment section below.

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