Question

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

Answer 1

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

$If \: n = 1 \implies S_{1} = 1^{2} + 13\times 1\\= 1+13 = 14$

$If \: n = 2 \implies S_{2} = 2^{2} + 13\times 2\\= 4 + 26 = 30$

$If \: n = 3 \implies S_{3} = 3^{2} + 13\times 3\\= 9 + 39 = 48$

$Now, First \:term (t_{1}) = S_{1} = 14$

$Second\:term (t_{2}) = S_{2} - S_{1} = 30 - 14 \\= 16$

$Third \:term (t_{3}) = S_{3} - S_{2}\\= 48 - 30\\= 18$

$t_{2} - t_{1} = 16 - 14 = 2 \: ---(1)$

$t_{3} - t_{2} = 18 - 16 = 2 \: ---(2)$

/* 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.,

$\blue { Yes }$

$\green { Sum \:of \:n \: terms \: in \:A.P \:is }$

$\green { in \:form \: n^{2} + 13n }$

•••♪