Question

In an arithmetic progression the sum of first n - term is
$\frac{5 {n}^{2} }{2} + \frac{3n}{2}$
find 17th term​

Answer 1

Step-by-step explanation:

go according to the pictures

Answer 2

ANSWER:

Given:

• Sum of firms n terms(S_n) = 5n²/2 + 3n/2

To Find:

• 17th term of the AP

Solution:

$\text{We are given that, sum of first n terms of an AP:}\\\\:\longrightarrow S_n=\dfrac{5n^2}{2}+\dfrac{3n}{2}\\\\\text{If we subtract S _n from S _{n+1} we will get the (n+1)th term.}\\\\\text{That is,}\\\\:\implies S_{n+1}-S_n=a_{n+1}\\\\\text{We need to find the 17th term. So,}\\\\:\implies (n+1)=17\\\\:\implies n=16\\\\\text{Hence,}\\\\:\implies S_{n+1}-S_n=a_{n+1}\\\\:\implies S_{16+1}-S_{16}=a_{16+1}\\\\:\implies S_{17}-S_{16}=a_{17}\\\\:\implies a_{17}=S_{17}-S_{16}\\\\:\implies a_{17}=\left(\dfrac{5(17)^2}{2}+\dfrac{3(17)}{2}\right)-\left(\dfrac{5(16)^2}{2}+\dfrac{3(16)}{2}\right)$

$:\implies a_{17}=\left(\dfrac{5(289)}{2}+\dfrac{51}{2}\right)-\left(\dfrac{5(256)}{2}+\dfrac{48}{2}\right)\\\\:\implies a_{17}=\dfrac{5(289)+51}{2}-\dfrac{5(256)+48)}{2}\\\\:\implies a_{17}=\dfrac{5(289)+51-5(256)-48}{2}\\\\:\implies a_{17}=\dfrac{5(289)-5(256)+51-48}{2}\\\\:\implies a_{17}=\dfrac{5(289-256)+3}{2}\\\\:\implies a_{17}=\dfrac{5(33)+3}{2}\\\\:\implies a_{17}=\dfrac{165+3}{2}\\\\:\implies a_{17}=\dfrac{168\!\!\!\!\!\!/^{\:\:\:84}}{2\!\!\!/_{\:1}}\\\\\bf{:\implies a_{17}=84}\\\\\text{\bf{Hence, the 17th term is 84}}$

Formula Used:

• S_(n+1) - S_(n) = a_(n+1)