Question

If the sum of the first n terms of an A.P. is given by Sn = 2n^2 + 7, then the nth term of the A.P. is (where '^' symbolise 'to the power') *

Answer 1

Answer:

Step-by-step explanation: Sn=2n² + 7

                                             let n=1

                                              first term =2+7=9

                                              now n=2 to get the sum of the first two terms

                                               2x2² + 7 =15

                                               this means that the first term and the second

                                                term added together is 15

                                                the first term =9 as calculated above

                                                 the second term would then be 15-9=6

                                                  common difference becomes 6-9= -3

                                                  the nth term becomes :a+(n-1)d

                                                  9+(n-1)-3

I really hope you understand

Answer 2

${\underline{\sf{Question}}}$

If the sum of the first n terms of an A.P. is given by Sn = 2n² + 7, then the nth term of the Ap is ?

${\underline{\sf{Formula's}}}$

Genral term of an AP

$\sf\:T_{n}=a+(n-1)d$

•For an AP

$\sf\:T_{n}=S_{n}-S_{n-1}$

•Common difference of An AP is given by

$\sf\:d=S_{2}-2S_{1}$

${\underline{\sf{Solution}}}$

Given: Sum of n terms, $\sf=S_{n}=2n{}^{2}+7$

Put n= 0

$\implies\:\sf\:S_{0}=7$

Put n = 1

$\implies\:\sf\:S_{1}=2\times1{}^{2}+7=9$

put n= 2

$\implies\:\sf\:S_{2}=8+7=15$

For an AP $\bf\:T_{n}=S_{n}-S_{n-1}$

First term,$\sf\:T_{1}=S_{1}-S_{1-1}$

$\implies\:\sf\:a_{1}=S_{1}-S_{0}$

$\sf\:a_{1}=9-7=2..(1)$

Second term,$\sf\:T_{2}=S_{2}-S_{2-1}$

$\sf\:a_{2}=S_{2}-S_{1}$

$\sf\:a_{2}=15-9=6$

Common difference, $\sf\:d=a_{2}-a_{1}$

$\sf\:d=6-2=4..(2)$

We know that,

Genral term of an Ap , $\sf\:T_{n}=a+(n-1)d$

$\implies\:\sf\:a_{n}=a+(n-1)d$

Now put the values of a and d from (1) and (2)

$\implies\:\sf\:a_{n}=2+(n-1)4$

$\implies\:\sf\:a_{n}=2+4n-4$

$\implies\:\sf\:a_{n}=4n-2$

$\rule{200}2$

More About Arithmetic progression:

Sum of n terms of an Ap is given by :

$\sf\:S_{n}=\dfrac{n}{2}[2a+(n-1)d]$