Answer 1

In the given problem, we are given first, second and last term of an A.P. We need to find its sum.

So, here

First term = a

Second term (a₂) = b

Last term (l) = 2a

Now, using the formula  a_n = a+ ( n - 1) d 

a₂ = a + (2 - 1)d

  b = a + d 

  d = b - a                 ......(1) 

Also,

a_n = a + (n-1) d

 2a = a + nd - d 

2a - a = nd - d 

`(a + d)/d = n `                     ...........(2) 

Further as we know,

`S_n = n/2 [ a + l]`

Substituting (2) in the above equation, we get

Using (1), we get

`S_n = ( a + (b - a) )/(2(b - a)) (3a) `

`S_n =b / ( 2 ( b-a) ) (3a)`

Thus,

`S_n = (3ab)/(2(b-a))`