Question

The sum of the first n terms of an A.P. is given by Sn = 2n2 + 5n , find the nth term of the A.P.

Answer 1

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

$\green{\tt{\therefore{A.P=7,11,15,...}}}$

$\green{\tt{\therefore{nth\:term=3+4n}}}$

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

$\green{\underline \bold{Fiven :}} \\ \tt: \implies Sum \: of \: n \: terms( s_{n}) = {2n}^{2} + 5n \\ \\ \red{\underline \bold{To \: find :}}\\ \tt: \implies A.P= ? \\ \\ \tt: \implies n th \:term \: of \: A.P= ?$

• According to given question :

$\tt \circ \: s_{n} = 2 {n}^{2} + 5n \\ \\ \tt \circ \: n = 1,2,3.....\\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies s_{1} = 2 \times {1}^{2} + 5 \times 1 \\ \\ \tt: \implies s_{1} =2 + 5 \\ \\ \green{\tt: \implies s_{1} =7 = a_{1}} \\ \\ \bold{For \: s_{2}} \\ \tt: \implies s_{2} = 2 \times {2}^{2} + 5\times 2 \\ \\ \tt: \implies s_{2} =8 + 10 \\ \\ \green{\tt: \implies s_{2} =18}\\ \\ \bold{for \: a_{2}} \\ \tt: \implies a_{2} = s_{2} - s_{1} \\ \\ \tt: \implies a_{2} =18 - 7 \\ \\ \tt: \implies a_{2} = 11 \\ \\ \bold{For \: common \: difference} \\ \tt: \implies Common \: difference = a_{2} - a_{1} \\ \\ \tt: \implies Common \: difference =11 - 7 \\ \\ \green{\tt: \implies Common \: difference =4} \\ \\ \tt\circ \: First \: term = 7 \\ \\ \tt \circ \: Common \: difference = 4\\ \\ \green{ \tt\therefore A.P = 7,11,15,....}$

$\bold{As \: we \: know \: that} \\ \tt: \implies a_{n} = a + (n - 1)d \\ \\ \tt: \implies a_{n} =7 + (n-1) \times 4 \\ \\ \tt: \implies a_{n} =7+ 4n-4 \\ \\ \green{\tt: \implies a_{n} =3+4n}$

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,

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

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

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

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

$\tt{ \green{A_{n} \: = 5 + n (14)\: = \: 5 + 14n}}.........(A)$