# CBSE Class 10 Answered

**Question 1) Find the sum of first 40 positive integers divisible by 6. Question 2) Find the sum of n terms of the series (4-1/n)+4-2/n)+(4-3/n)+........ Question 3) If the sum of the first n terms of an AP is 1/2(3n²+7n), then find its nth term .Hence write the 20 th term .**

Asked by arindeep.singh | 22 Jul, 2020, 06:18: PM

Expert Answer

(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

S_{40}= 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 (3n^{2}+7n)

Take n=1, S_{1} = 1/2 (3+7) = 5

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

We know, S_{1}=a_{1}=5

S_{2}=a_{1}+a_{2}=13

S_{2}-S_{1}=a_{1}+a_{2}-a_{1}

13-5=a_{2}

Therefore, a_{2}=8

Also, d = a_{2}-a_{1}

d = 8-5 = 3

n^{th} term of AP = a_{n} = 5+(n-1)3

a_{n}= 2+3n

Therefore 20^{th} term is

a_{20}= 2+3(20)=62

Hence 20th term is 62.

Given: S_{n }= 1/2 (3n^{2}+7n)

Take n=1, S_{1} = 1/2 (3+7) = 5

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

We know, S_{1}=a_{1}=5

S_{2}=a_{1}+a_{2}=13

S_{2}-S_{1}=a_{1}+a_{2}-a_{1}

13-5=a_{2}

Therefore, a_{2}=8

Also, d = a_{2}-a_{1}

d = 8-5 = 3

n^{th} term of AP = a_{n} = 5+(n-1)3

a_{n}= 2+3n

Therefore 20^{th} term is

a_{20}= 2+3(20)=62

Hence 20th term is 62.

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

S_{40}= 20[2(6) +(40-1)6] =20[12+234] =4920

Answered by Renu Varma | 23 Jul, 2020, 01:00: PM

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