Question

The sum of first n terms of an A.P. is given by Sn = 3n2 + 2n. Determine the A.P. and its 15th term.

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{A.P=5,11,17,...}}}$

$\green{\tt{\therefore{15th\:term=89}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Fiven :}} \\ \tt: \implies Sum \: of \: n \: terms( s_{n}) = {3n}^{2} + 2n \\ \\ \red{\underline \bold{To \: find :}}\\ \tt: \implies A.P= ? \\ \\ \tt: \implies 15 th \:term \: of \: A.P= ?$

• According to given question :

$\tt \circ \: s_{n} = 3 {n}^{2} + 2n \\ \\ \tt \circ \: n = 1,2,3.....\\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies s_{1} = 3 \times {1}^{2} + 2 \times 1 \\ \\ \tt: \implies s_{1} =3 + 2 \\ \\ \green{\tt: \implies s_{1} =5 = a_{1}} \\ \\ \bold{For \: s_{2}} \\ \tt: \implies s_{2} = 3 \times {2}^{2} + 2 \times 2 \\ \\ \tt: \implies s_{2} =12 + 4 \\ \\ \green{\tt: \implies s_{2} =16}\\ \\ \bold{for \: a_{2}} \\ \tt: \implies a_{2} = s_{2} - s_{1} \\ \\ \tt: \implies a_{2} = 16 - 5 \\ \\ \tt: \implies a_{2} = 11 \\ \\ \bold{For \: common \: difference} \\ \tt: \implies Common \: difference = a_{2} - a_{1} \\ \\ \tt: \implies Common \: difference =11 - 5 \\ \\ \green{\tt: \implies Common \: difference =6} \\ \\ \tt\circ \: First \: term = 5 \\ \\ \tt \circ \: Common \: difference = 6\\ \\ \green{ \tt\therefore A.P = 5,11,17,....}$

$\bold{As \: we \: know \: that} \\ \tt: \implies a_{n} = a + (n - 1)d \\ \\ \tt: \implies a_{15} =5 + (15 - 1) \times 6 \\ \\ \tt: \implies a_{15} =5 + 14 \times 6 \\ \\ \green{\tt: \implies a_{15} =89}$

Answer 2

$</p><p>\tt{\huge{\pink{Hello!!! }}}$

$</p><p>\tt{\red{Given \: - }}$

$\tt{ \orange{S_{n} \: = 3 {n}^{2} + 2n \: }}$

Let there be :

• a is the first term.

• d is the common difference between any two successive terms.

Hence,

$\tt{ \purple{ \frac{n}{2} (2a \: + \: (n - 1)d \: ) \: = n( 3 n + 2 \: )}}$

$\tt{ \red{S_{1} \: = 3 {1}^{2} + 2 \: = \: 5}}$

$\tt{ \pink{S_{2} \: = 3 {2}^{2} + 4\: = \: 16}}$

$\tt{\blue{\implies {d = S_{2} - S_{1} = 9 }}}$

$\tt{ \green{A_{15} \: = 5 + 6 (14)\: = \: 89}}.........(A1)$

$\tt{\red{The \: AP \: Is \: 5, \: 16 , \: 25, \: ..........(A2)}}$