Question

if sum of n terms of an AP is 2n²+5, then prove that an=4n+3. fast please

Answer 1

Heya.....

Here's your answer....

Let's take a general A.P. where the starting term is "a" and the common difference is "d".

Let S(n) denote the sum of N terms.

S(n) = (n(2a + (n-1)d))/2

= an + (n^2×d)/2 - (nd/2)

Group all alike terms

S(n) = n×(a-d/2) + n^2×(d/2)

Assuming you miswrote 2n^2 + 5n as just 5 which I think it should be, but I could be wrong...

d/2 = 2

d = 4

a - d/2 = 5

a = 5 + 2

a = 7

A(n) = a + (n-1)d

= 7 + (n-1)4

= 7 + 4n - 4

= 4n + 3

Thanks...!!!

XD

Sorry baby 'wink'