MCQ

If the first, second and last term of an A.P. are *a*, *b* and 2*a* respectively, its sum is

#### Options

- \[\frac{ab}{2(b - a)}\]
- \[\frac{ab}{(b - a)}\]
- \[\frac{3ab}{2(b - a)}\]
none of these

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#### Solution

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 (*a*_{2}) = b

Last term (*l*__)__ = 2*a*

Now, using the formula a_{n} = a+ ( n - 1) d

a_{2} = 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))`

Concept: Sum of First n Terms of an AP

Is there an error in this question or solution?

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