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 | 22nd 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

 

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= 1/2 (3n2+7n)

 Take n=1, S1 = 1/2 (3+7) = 5 

Take n=2, S= 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

nth term of AP = an = 5+(n-1)3

an= 2+3n

Therefore 20th term is 

a20= 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

 

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

 

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