Question

Please help with this

Answer 1

`S_(25)=555`

Explanation:

Given that,

In an AP, the 6th term is 39 i.e.

`a_6=a+5d\\\\39=a+5d\ ....(1)`

In the same AP, the 19th term is 7.8 i.e.

`a_(19)=a+18d\\\\7.8=a+18d\ ......(2)`

Subtract equation (1) from (2).

7.8 - 39 = a+18d - (a+5d)

-31.2 = a +18d-a-5d

-31.2 = 13d

d = -2.4

Put the value of d in equation (!).

39 = a+5(-2.4)

39 = a- 12

a = 39+12

a = 51

The sum of n terms of an AP is given by :

`S_n=(n)/(2)[2a+(n-1)d]`

Put n = 25 and respected values,

`S_(25)=(25)/(2)[2(51)+24(-2.4)]\\\\=555`

Hence, the sum of first 25 terms of the AP is equal to 555.