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.**

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. **

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