Question

Find the sum of the first 30 terms of the sequence below an=3n+2

Answer 1

First off, let's find the 1st term's value, and the 30th term's value,

$\bf a1=3(1)+2\implies a1=5\qquad \qquad \quad a30=3(30)+2\implies a30=92\\\\ -------------------------------\\\\ ~~~~~~~\textit{ Sum of an arithmetic sequence}\\\\ S_n=\cfrac{n(a1+an)}{2}~ \begin{cases} n=n^{th}\ term\\ a1=\textit{first term's value}\\ ----------\\ a1=5\\ a30=92\\ n=30 \end{cases} \implies S_{30}=\cfrac{30(5+92)}{2} \\\\\\ S_{30}=15(97)$

Answer 2

Answer:

$S_{30}$= 1455.

Step-by-step explanation:

Given : an=3n+2.

To find : find the sum of the first 30 terms of the sequence.

Solution : We have given $a_{n}$ = 3n + 2.

For first term n = 1

 $a_{1}$ = 3 (1) +2.

$a_{1}$ = 5

For last term n = 30

$a_{30}$ = 3(30) +2

$a_{30}$ = 90 +2

$a_{30}$ = 92.

Then sum of first 30 terms

$S_{30} =\frac{n(First\ term+last\ term)}{2}$.

$S_{30} =\frac{30(5+92)}{2}$.

$S_{30} =\frac{30(97)}{2}$.

$S_{30}$= 15 *97.

$S_{30}$= 1455.

Therefore, $S_{30}$= 1455.