Question

S = n/2(2a+(n-1)d) make n as subject​

Answer 1

Consider an AP consisting n terms.

First Term = a

Common Difference = d

nth term = an

Sum of n term is given as S=n2[2a+(n−1).d]

Proof:

Consider an AP : a, a+d, a+2d, …………., a+(n-1).d

Sum of first n terms = a + (a+d) + (a+2d) + ………. + [a+ (n-1).d] ——————-(i)

Writing the terms in reverse order, we have: S = [a+(n-1).d] + [a+(n-2).d] + [a+(n-3).d] + ……. (a) ———–(ii)

Adding both the equations term wise, we have

2S = [2a + (n-1).d] + [2a + (n-1).d] + [2a + (n-1).d] + …………. + [2a + (n-1).d] (n-terms)

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

S=n2[2a+(n−1).d]

Answer 2

Answer:

n the term which we take

Step-by-step explanation:

it is sum of n terms formulae