Question

What is the sum of a 58th arithmetic sequence where the first term is 6 and the last term is 405

Answer 1

Formula to find the sum of nth terms of an arithmetic sequence is:

$S_{n} =\frac{n}{2} (a_{1} +a_{n})$

Where, $S_{n}$ = sum of nth terms.

$a_{1}$ = first term.

$a_{n}$ = last term.

Now we need to find, sum of a 58th arithmetic sequence where the first term is 6 and the last term is 405 .

So, plug in n = 58, $a_{1}$ = 6 and $a_{n}$ = 405 in the above formula.

$S_{58} =\frac{58}{2} (6+405)$

= $\frac{58}{2} (411)$

=$\frac{23838}{2}$

= 11919

So, the sum of 58th terms is 11919.

Hope this helps you!.