# The 3rd term of an AP is 24 and the 5th term is 36. What is the sum of the first 20 terms? Let a1, a2, a3, a4, a5, ..., a20 are terms of an A.P

Given that:

a3 = 24

a5 36

We need to find the sum of the first 20 terms.

First we need to find the value of a1 and the common difference r.

We know that:

a3= a1+...

Let a1, a2, a3, a4, a5, ..., a20 are terms of an A.P

Given that:

a3 = 24

a5 36

We need to find the sum of the first 20 terms.

First we need to find the value of a1 and the common difference r.

We know that:

a3= a1+ 2r = 24 .............(1)

a5= a1+ 4r = 36 ..............(2)

We will subtract (1) from (2) .

==>  2r = 12 ==> r= 6  ==> a1= 24- 12 = 12

Now we know that the sum of n terms of A.P is given by :

Sn = (n/2)*(2a1+ (n-1)*r )

==> S20 = (20/2) ( 2*12 + 19*6) = 10* 138 = 1380.

Then the sum of the first 20 terms is 1380.

Approved by eNotes Editorial Team The nth term of an AP can be written as Tn = a + (n-1)d.

T3 = a + 2d = 24

T5 = a + 4d = 36

T5 – T3 = 2d = 12

=> d = 6

a = 24 – 2d = 12

The sum of the first 20 terms is (T1 + T20)*(20/2)

=> (12 + 12 + 19*6)*10

=> 138*10

=> 1380

The required sum of the first 20 terms is 1380.

Approved by eNotes Editorial Team