The sum of n terms of the AP √3, √8, √18, √32, is

S = √2+√8+√18+√32+……………… n terms.

or, S = √2 + 2√2 +3√2 +4√2.+………………….+ n terms.

This is an A.P. in which a = √2. , d = √2.

Sn = n/2.[2.a +(n-1).d].

or, Sn = n/2.[ 2√2 +(n-1).√2].

or, Sn = (n/2).√2.[ 2 +n-1].

or, Sn = n.(n+1)/√2. Answer.

Given AP :√2,√8,√18,√32,...

First term = √2

Common difference = √8-√2

⇛ √(2×2×2)-√2

⇛ 2√2-√2

⇛√2

We know that

Sum of first n terms of an AP

⇛ Sn = (n/2)[2a+(n-1)d]

⇛ Sn = (n/2)[2√2+(n-1)(√2)]

⇛ Sn = (n/2)[2√2+√2 n -√2]

⇛ Sn = (n/2)[√2 n +√2)

⇛ Sn = (n/2)×(√2)(n+1)

⇛Sn = (n/√2)(n+1)

⇛ Sn = n(n+1)/√2

Additional comment:

nth term of an AP = a+(n-1)d

Sum of first n terms of an AP

Sn = (n/2)[2a+(n-1)d]

Sum of the first n terms = Sn =

(n/2)(a+an)

If a,b,c are the three consecutive terms in an AP then b = (a+c)/2

also read similar questions: Find the sum of last ten terms of the AP : 8, 10, 12,, 126.

The sum of the 8th & 4th of an Ap is 24 and sum of and sum of 6th & 10th term is 44 Find the Ap

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