Question

find the sum of 51 terms of an ap whose second and third terms are 14 and 18 respectuvely.

Answer 1

2nd Term = 14

So, a+d = 14 ------------- Eq. 1

3rd Term = 18

So, a+2d = 18 -----------Eq. 2

Subtract Eq 1 from Eq.2

Then , a + 2d = 18

- a - d = -14

-------------------------

d = 4

Then substitute the value of d in Eq. 1

Then, a + 4=14

So, a = 10

Sum = n/2 ( 2a ( n-1 ) d)

=51/2 ( 2*10 * 50*4)

= 51/2 ( 20 * 200)

= 51/2 * 4000

= 102000

So, the Sum of 51 terms is 102000.

Thanks

Regards

Aditya Suryawanshi

Answer 2

the correct answer of sum is 3060