Question

the sum of first n terms of an A.P. is 3n*n+4n. find the 25th term of the A.P.

Answer 1

since
$sn \: = 3 {n}^{2} + 4n \\ then \: s1 = 3 {(1)}^{2} + 4(1) \\ s2 = 3( {2}^{2} ) + 4(2) \\ = 3(4) + 8 \\ = 12 + 8 = 20 \\ and \: s2 = \frac{2}{2} (2a + nd - d) \\ 2a + 4d - d = 20 \\ 2(7) + 3d = 20 \\ 3d = 20 - 14 \\ 3d = 6 \\ d = \frac{6}{3} \\ d = 2 \: and \: a = 7 \\ then \: tn \: a + (n - 1)d$
t25= 7+(24)2
7+98 = 105