Can the sum of n terms in AP is n square+13n

1.	Can the sum of n terms in AP is n square+13n
Answer» $Given \:Sum \:of \:n\:terms \:in \:A.P:\\ (S_{n}) = n^{<klux>2</klux>} + 13n$ $If \: n = <klux>1</klux> \implies S_{1} = 1^{2} + 13\times 1\\= 1+13 = 14$ $If \: n = 2 \implies S_{2} = 2^{2} + 13\times 2\\= 4 + 26 = 30$ $If \: n = 3 \implies S_{3} = 3^{2} + 13\times 3\\= 9 + 39 = 48$ $Now, First \:term (t_{1}) = S_{1} = 14$ $Second\:term (t_{2}) = S_{2} - S_{1} = 30 - 14 \\= 16$ $Third \:term (t_{3}) = S_{3} - S_{2}\\= 48 - 30\\= 18$ $t_{2} - t_{1} = 16 - 14 = 2 \: ---(1)$ $t_{3} - t_{2} = 18 - 16 = 2 \: ---(2)$ /* From (1) and (2) ,*/ $t_{2} - t_{1} =t_{3} - t_{2} = 2$ $\pink { (Difference \:of \: consecutive \:terms }$ $\pink { are \: equal ) }$ $14,16,18, \cdot \cdot \cdot , is \:an \:A.P$ THEREFORE., $\blue { Yes }$ $\green { Sum \:of \:n \: terms \: in \:A.P \:is }$ $\green { in \:form \: n^{2} + 13n }$ •••♪

Answer»

$Given \:Sum \:of \:n\:terms \:in \:A.P:\\ (S_{n}) = n^{<klux>2</klux>} + 13n$