Question

Find the sum of first 25 terms of the AP 3,9/2,6,15/2,....

Answer 1

Let:

a=first term

d=common difference

AP=3,9/2,6.........

d=9/2-3

=9-6/2

=3/2=1.5

Given:a=3

So

Sn=n/2*[2a+(n-1)d]

S25=25/2*[6+(25-1)1.5]

=25/2[6+24*1.5]

=25/2*[6+36]

=25/2*42

=25*21

=525.

That's the answer: 525.

Hope it helps.

Answer 2

Given,

A.P. 3, 9/2, 6, 15/2

To Find,

The sum of the first 25 terms.

Solution,

We can solve the question as follows:

It is given that we have to find the sum of the first 25 terms of the A.P. 3, 9/2, 6, 15/2.

In the A.P. the first term is 3. The common difference, d is equal to the difference between any two consecutive terms. Therefore,

$d = \frac{9}{2} - 3$

  $= \frac{9-6}{2}$

  $= \frac{3}{2}$

Now,

The formula for the sum of the first n terms of an A.P. is given as:

$S_{n} = \frac{n}{2} (2a + (n - 1)d)$

Where,

$n = Number\: of\: terms$

$a = First\: term$

$d = Common\: difference$

To find the sum of the first 25 terms, we will substitute the given values in the above formula.

$S_{25} = \frac{25}{2} (2*3 + (25 - 1)\frac{3}{2} )$

     $= \frac{25}{2} (6 + 24*\frac{3}{2} )$

     $= \frac{25}{2} ( 6 + 12*3)$

     $= \frac{25}{2} ( 6 + 36)$

     $= \frac{25}{2} *42$

     $= 25*21$

     $= 525$

Hence, the sum of the first 25 terms is 525.