Question

the nth term of an arithmetic progression is given by Tn=7n+1. find the sum of the first 30 terms of arithmetic prigression.

Answer 1

Given that nth term of an arithmetic progression is Tn=7n+1

where n can be n=1,2,3,..

plugging n=1 gives first term

T1= 7*1+1=7+1=8

plugging n=2 gives second term

T2= 7*2+1=14+1=15

Common difference "d" can be found by difference of both terms

d=T2- T1= 15-8=7

Now we need to find sum of first 30 terms of arithmetic progression.

So we will use formula

$S_n=\frac{n}{2}(2a+(n-1)d)$

where n=30, a=first term = 8, d=common difference = 7

plug those values

$S_{30}=\frac{30}{2}(2*8+(30-1)*7)$

$S_{30}=15(16+(29)*7)$

$S_{30}=15(16+203)$

$S_{30}=15(219)$

$S_{30}=3285$

Hence final answer is 3285.

Answer 2

Answer:

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