Question

it the 10th term of an AP is 47 and its first term is 2,find the sum of its first 15 term

Answer 1

an= a + (n-1)d
47=2+9d
5=d
sn = n/2(2a+(n-1)d)
    =15/2(4+14*5)
    =555

Answer 2

Hey there!

Let the first term be a, and common difference be d.

Given,
Tenth term of the A. P = 47

a + 9d = 47 .

Also,
First term ( a) = 2 .

So,
2 + 9d = 47
9d = 45
d = 5 .

Now,
Sum of first n terms = n/2 [ 2a + ( n - 1 )d ]

Sum of first 15 terms = 15/2 [ 2*2 + (15-1)5 ]

= 15/2 ( 4 + 14 * 5 )

= 15/2 ( 4 + 70 )

= 15 ( 37 )

= 555 .

Therefore , Sum of first 15 terms is 555