CBSE Class 10 Answered

(1)

We know that the first 40 positive integers divisible by 6 are 6,12,18,....

This is an AP with a = 6 and d = 6

 

S₄₀= 20[2(6) +(40-1)6] =20[12+234] =4920

(2)

(4-1/n)+(4-2/n)+(4-3/n)+.....+(4-n/n)

= (4+4+...+upto n terms) - 1/n(1+2+3+...+n)

= 4n - 1/n[n(n+1)/2]

= 4n - (n+1)/2

= (8n - n - 1)/2

= (7n-1)/2

 

(3)

Given: S_n = 1/2 (3n²+7n)

 Take n=1, S₁ = 1/2 (3+7) = 5 

Take n=2, S₂ = 1/2 (3x4 + 7x2) = 26/2 = 13

We know, S₁=a₁=5

S₂=a₁+a₂=13

S₂-S₁=a₁+a₂-a₁

13-5=a₂

Therefore, a₂=8

 Also, d = a₂-a₁

d = 8-5 = 3

n^th term of AP = a_n = 5+(n-1)3

a_n= 2+3n

Therefore 20^th term is 

a₂₀= 2+3(20)=62

Hence 20th term is 62.