Question

Find the sum of 3a,a,-a,.....to ,a terms

Answer 1

The AP is as follows :-

$\boxed{\bold{\red{3a,a,(-a),....... a\:terms}}}$

$\rule{200}{2}$

Here,

A=3a

d = (-2a)

$\rule{200}{2}$

We know that the sum of n terms of AP is always equal to :-

$\boxed{\boxed{\bold{\sf{\blue{s_n = \dfrac{n}{2}(2a+(n-1)d)}}}}}$

$\rule{200}{2}$

Applying the same Formula here also :-

Sum of a terms of the AP will be as follows :-

$\frac{a}{2}(2(3a) + (a - 1)( - 2)) \\ \\ \\ = \frac{a}{2}(6a - 2a + 2) \\ \\ \\ = \frac{a}{2}(2(2a + 1)) \\ \\ \\ = a(2a + 1) \\ \\ \\ = 2 {a}^{2} + a$

$\rule{200}{2}$

Answer 2

Answer:

Step-by-step explanation:

Given,

3a,a,_a.......a terms

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

Sa=a/2(6a-2a+1)

Sa=a/2(2(2a+1)

Sa=a(2a+1)

Sa=2a^2+a