Question

the sum of the first n terms of an AP is 5n^2-3n. find the AP and hence find its 12 th term

Answer 1

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

$\green{\tt{\therefore{12th\:term=112}}}$

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

$\green{\underline \bold{Given :}} \\ \tt: \implies Sum \: of \: first \: n \: terms(s_{n})= 5 {n}^{2} - 3n \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies 12th \: term \: of \: A.P = ?$

• According to given question :

$\tt \circ \: s_{n} =5 {n}^{2} - 3n \\ \\ \tt \because \: n = 1,2,3,.. \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies s_{1} = 5 \times {1}^{2} - 3 \times 1 \\ \\ \tt: \implies s_{1} =5 - 3 \\ \\ \green{\tt: \implies s_{1} =2 = a_{1} } \\ \\ \bold{For \: s_{2}(n = 2)} \\ \tt: \implies s_{2} = 5 \times {2}^{2} - 3 \times 2 \\ \\ \tt: \implies s_{2} =5 \times 4 - 6 \\ \\ \tt: \implies s_{2} =20 - 6 \\ \\ \green{\tt: \implies s_{2} =14}\\ \\ \bold{For \: Second \: term} \\ \tt: \implies a_{2} = s_{2} - s_{1} \\ \\ \tt: \implies a_{2} = 14 - 2 \\ \\ \green{\tt: \implies a_{2} =12} \\ \\ \bold{For \:Common \:Difference : } \\ \tt: \implies d= a_{2} - a_{1} \\ \\ \tt: \implies d=12 - 2 \\ \\ \green{\tt: \implies d=10} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies a_{n}=a + (n - 1)d \\ \\ \tt: \implies a_{12}=2 + (12 - 1) \times 10 \\ \\ \tt: \implies a_{12}=2 + 110 \\ \\ \green{\tt: \implies a_{12}=112}$

Answer 2

QUESTION :

The sum of the first n terms of an AP is 5n^2-3n. find the AP and hence find its 12 th term.

SOLUTION :

• According to given question :

$\begin{lgathered}\tt \circ \: s_{n} =5 {n}^{2} - 3n \\ \\ \tt \because \: n = 1,2,3,.. \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies s_{1} = 5 \times {1}^{2} - 3 \times 1 \\ \\ \tt: \implies s_{1} =5 - 3 \\ \\ \green{\tt: \implies s_{1} =2 = a_{1} } \\ \\ \bold{For \: s_{2}(n = 2)} \\ \tt: \implies s_{2} = 5 \times {2}^{2} - 3 \times 2 \\ \\ \tt: \implies s_{2} =5 \times 4 - 6 \\ \\ \tt: \implies s_{2} =20 - 6 \\ \\ \green{\tt: \implies s_{2} =14}\\ \\ \bold{For \: Second \: term} \\ \tt: \implies a_{2} = s_{2} - s_{1} \\ \\ \tt: \implies a_{2} = 14 - 2 \\ \\ \green{\tt: \implies a_{2} =12} \\ \\ \bold{For \:Common \:Difference : } \\ \tt: \implies d= a_{2} - a_{1} \\ \\ \tt: \implies d=12 - 2 \\ \\ \green{\tt: \implies d=10} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies a_{n}=a + (n - 1)d \\ \\ \tt: \implies a_{12}=2 + (12 - 1) \times 10 \\ \\ \tt: \implies a_{12}=2 + 110 \\ \\ \orange{\tt: \implies a_{12}=112}\end{lgathered}$