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

Question:

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

(a) $\frac{a b}{2(b-a)}$

(b) $\frac{a b}{(b-a)}$

(c) $\frac{3 a b}{2(b-a)}$

 

(d) none of these

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 (a2) = b

Last term (l) = 2a

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

$a_{2}=a+(2-1) d$

$b=a+d$

 

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

Also,

$a_{n}=a+(n-1) d$

$2 a=a+n d-d$

$2 a-a=n d-d$

$\frac{a+d}{d}=n$ ...........(2)

Further as we know,

$S_{n}=\frac{n}{2}[a+l]$

Substituting (2) in the above equation, we get

Using (1), we get

$S_{n}=\frac{a+(b-a)}{2(b-a)}(3 a)$

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

Thus,

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

Therefore, the correct option is (c).

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