CG Board Class 10 Maths Chapter 4 Solutions: Arithmetic Progression (समांतर श्रेणी) (2026-27)

(i) First term and common difference of given arithmetic progression are: 3/2, 1/2, -1/2, -3/2, …**

• (ii) If first term is -2 and common difference is -2 for an arithmetic progression, then fourth term is:
• (iii) 15th term in the arithmetic progression 7, 13, 19, … is:
• (iv) First term is 4 and common difference is -4 of an arithmetic progression, then nth term is:
• (v) 78 is which term of arithmetic progression 3, 8, 13, 18, …?

• (a) a, a², a³, a⁴, …

a₂ – a₁ = a² – a; a₃ – a₂ = a³ – a². Since a₂ – a₁ ≠ a₃ – a₂, it is not an AP.

• (b) √2, √8, √18, √32, …

Simplifying terms: √2, 2√2, 3√2, 4√2, …

a₂ – a₁ = 2√2 – √2 = √2; a₃ – a₂ = 3√2 – 2√2 = √2.

Since the common difference d = √2 is constant, it is an AP.

• (c) 1², 3², 5², 7², 9², …

Terms are 1, 9, 25, 49, 81, ….

a₂ – a₁ = 9 – 1 = 8; a₃ – a₂ = 25 – 9 = 16. Differences are not equal, so it is not an AP.

• (d) 0, -4, -8, -12, …

a₂ – a₁ = -4 – 0 = -4; a₃ – a₂ = -8 – (-4) = -4.

Common difference d = -4 is constant, so it is an AP.

• (e) 16, 181/2, 201/2, 23, …

a₂ – a₁ = 18.5 – 16 = 2.5; a₃ – a₂ = 20.5 – 18.5 = 2.0.

Differences are not equal (2.5 ≠ 2.0), so it is not an AP.

27**

1070**

(3n – 2)/(9)**

1000 – 50m**

31 terms*

10th term*

99**

10 rows*

n = 10**

Yes, 27th term*

178**

2m + 1**

65th term*

13, -8, -3**

300**

208**

1 )*

• Proof /

a + (p – 1)d = 1/q — (1), a + (q – 1)d = 1/p — (2)

Subtracting (2) from (1): (p – q)d = (p – q)/(pq) ⇒ d = 1/pq

Substitute d: a = 1/pq

a_pq = a + (pq – 1)d = 1/pq + (pq – 1)1/pq = 1

• Proof /

a + (p – 1)d = q — (1), a + (q – 1)d = p — (2)

Subtracting (2) from (1): (p – q)d = -(p – q) ⇒ d = -1

From (1): a = p + q – 1

a_p+q = a + (p + q – 1)d = (p + q – 1) + (p + q – 1)(-1) = 0

+ b(r – p) + c(p – q) = 0)*

• Proof /

Let A be first term and D be common difference.

a = A + (p – 1)D, b = A + (q – 1)D, c = A + (r – 1)D

a(q – r) + b(r – p) + c(p – q) = A[(q – r) + (r – p) + (p – q)] + D[(p – 1)(q – r) + (q – 1)(r – p) + (r – 1)(p – q)] = A(0) + D(0) = 0

n = 13**

0**

(x² + y²)/(2)**

5 and 9*

3 and 9*

4, -2, 0, 2, 4, 6, 8, 10**

11, 8, 5, 2, -1, -4, -7**

• Proof /

(a_p + a_q)/(2) = (a_r + a_s)/(2) ⇒ [a + (p – 1)d] + [a + (q – 1)d] = [a + (r – 1)d] + [a + (s – 1)d]

2a + (p + q – 2)d = 2a + (r + s – 2)d ⇒ p + q = r + s

n = 9*

(i) 9, 12, 15, … up to 16 terms:**

a = 9, d = 3, n = 16

S₁₆ = 16/2[2(9) + (16 – 1)3] = 8[18 + 45] = 8 × 63 = 504

504**

• (ii) 8, 3, -2, … up to 22 terms:

a = 8, d = -5, n = 22

S₂₂ = 22/2[2(8) + (22 – 1)(-5)] = 11 = 11 × (-89) = -979

-979**

• (iii) 0.6, 1.7, 2.8, … up to 100 terms:

a = 0.6, d = 1.1, n = 100

S₁₀₀ = 100/2[2(0.6) + (100 – 1)1.1] = 50[1.2 + 108.9] = 5505

5505**

• (iv) 1/15, 1/12, 1/10, … up to 11 terms:

a = 1/15, d = 1/12 – 1/15 = 1/60, n = 11

S₁₁ = 11/2[2/15 + 10/60] = 11/2 × 9/30 = 33/20

33/20**

• (v) n² + 1, n² – 1, n² – 3, … up to 20 terms:

a = n² + 1, d = -2, n = 20

S₂₀ = 20/2[2(n² + 1) + 19(-2)] = 10[2n² – 36] = 20(n² – 18)

20(n² – 18) or 210(2n – 17) as per key**

• (vi) (1 – 1/n), (1 – 2/n), (1 – 3/n), … up to n terms:

a = 1 – 1/n, d = -1/n

S_n = n/2[2(1 – 1/n) + (n – 1)(-1/n)] = n/2[(n – 1)/(n)] = (n – 1)/(2)

(n – 1)/(2)**

23 terms*

4 or 13 terms*

n = 6, d = 2**

n = 38, Sum = 6973**

1683**

625**

5610**

n²**

/(2(b – a)) )*

• Proof /

First term = a, d = b – a, last term l = 2a

a + (n – 1)(b – a) = 2a ⇒ n – 1 = (a)/(b – a) ⇒ n = (b)/(b – a)

S_n = n/2(a + l) = ((b)/(b – a))/(2)(a + 2a) = (3ab)/(2(b – a))

33**

672**

+ b/q(r – p) + c/r(p – q) = 0)*

• Proof /

a/p = A + 1/2(p – 1)D, b/q = A + 1/2(q – 1)D, c/r = A + 1/2(r – 1)D

Multiplying by (q – r), (r – p), (p – q) and adding gives 0.

• Proof /

s₁ = (n(n + 1))/(2), s₂ = n², s₃ = (n(3n – 1))/(2)

s₁ + s₃ = (n(n + 1 + 3n – 1))/(2) = (4n²)/(2) = 2n² = 2s₂

Prizes: Rs. 130, 120, 110, 100, 90, 80, 70**

16 rows, 5 logs in the top row*

370 meters*