Question

If a,b,and 2a in ap show that 3ab/2(b-a)​

Answer 1

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

Explanation:

The correct question is: If the first, second and last term of an AP are a,b and 2a respectively, then show that the sum of all terms of an AP is 3ab/2(b-a)​.

Firstly, as we know that the nth term of an A.P. is given by the following formula;

`a_n=a+(n-1)d` , where a = first term of AP, d = common difference, n = number of terms in an AP and
`a_n` = last term

Since it is given that the first, second and last term of an AP are a,b and 2a respectively, that means;

first term = a

d = second term - first term = b - a

`a_n` = 2a

So,
`a_n=a+(n-1)d`

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

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

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

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

`(a)/(b-a) +1= n`

`(a+(b-a))/(b-a) = n`

`n=(b)/(b-a)` ------------- [equation 1]

Now, the formula for the sum to n terms of an AP when the last term is given to us is;

`S_n = (n)/(2)[\text{first term} + \text{last term}]`

`S_n = (b)/(2* (b-a))[a +2a]` {using equation 1}

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

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

Hence proved.