Question

Sum of the first 14 terms of an A.P. is 1505 and its first term is 10. Find its 25th term.

Answer 1

Sum of n terms = Sn = n/2 [ 2a + (n-1) d ]
where a = first term d = common difference 

therefore, S14 = 14/2 [2*10 + (14-1) d ] = 1505

solve for d, we get, 215 = 20 + 13d
hence, d = 15. 

now, an = a + (n-1) d
therefore, a25 = 10 + (25-1) 15
a25 = 10 + 24 * 15 = 370. 

Answer 2

Given,

Sum of the first 14 terms of an A.P. is 1505 and its first term is 10.

To find,

Find its 25th term.

Solution,

We know that,

Sum of n terms = $S_{n}$ = n/2 [ 2a + (n-1) d ]

where a = first term d = common difference

So,

$S_{14}$ = 14/2 [(2 x 10) + (14-1) d ] = 1505

By solving for d, we get

215 = 20 + 13d

13d = 195

d = 195/13 = 15

Now,

$n^{th}$ term = $a_{n}$ = a + (n-1) d

So,

$a_{25}$ = 10 + (25-1) 15

     = 10 + 24 * 15 = 370

Hence, 370 is the 25th term of the A.P.