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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

### (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: Sn = 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 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 | 23 Jul, 2020, 01:00: PM

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