Question

first ,second and last term of an ap are a,b,2a. Then find the sum​

Answer 1

Answer:

$S_{n} = \dfrac{3ab}{2(b-a)}$

Step-by-step explanation:

Given that-

The first term of A.P = a

Second term of A.P = b

Last term of A.P = 2a

We know that-

Difference of A.P = second term - first term

d = b - a

We know that-

The last term of A.P is-

$T_n} = a+(n-1)d$

2a = a + (n - 1).(b - a)

2a - a = (n - 1)(b - a)

a = (n - 1)(b - a)

$n - 1 = \dfrac{a}{b - a}$

$n = \dfrac{a}{b-a} + 1$

$n = \dfrac{a+b-a}{b-a}$

$n = \dfrac{b}{b-a}$

So, the sum of n terms-

$S_{n} = \dfrac{n}{2}[T_{1} - T_{n}]$

$= \dfrac{b}{2(b-a)}[a+2a]$

$S_{n} = \dfrac{b}{2(b-a)} \times3a$

$S_{n} = \dfrac{3ab}{2(b-a)}$