An algebra problem by Keerthi Reddy

First find the value for n = 25 and then subtract value of n = 24 from it.

Let $S_n = \dfrac 12 (3n^2 + 13 n)$

$a_{25} = S_{25} - S_{24} = \dfrac 12 \begin{pmatrix} 3 ( 25^2 - 24^2 ) + 13(25 -24) \end{pmatrix} = \dfrac 12 (3 \times 49 + 13 \times 1 ) = \boxed {80}$

Substitute n =1: the first term is 8.

Substitute n = 2: you get 19. So the second term is 19-8 = 11.

The difference is 3: the 25th term = 8 + (24 *3) = 80

$S_{25} = T_1 + T_2 + T_3 + \ldots + T_{23} + T_{24} + T_{25}\\ S_{25} = \left( T_1 + T_2 + T_3 + \ldots + T_{23} + T_{24} \right) + T_{25}\\ S_{25} = S_{24} + T_{25}\\ T_{25} = S_{25} - S_{24}\\ =\dfrac{3(25)^2+13(25)}{2} - \dfrac{3(24^2) + 13(24)}{2}\\ =\dfrac{3(25^2-24^2) + 13(25-24)}{2}\\ =\dfrac{3(25-24)(25+24) + 13}{2}\\ =\dfrac{147+13}{2}\\ =\boxed{80}$

Find for n = 25 and then subtract n = 24 from it.