Question

What is the sum of a 56-term arithmetic sequence where the first term is 6 and the last term is 391?

Answer 1

$S_n=\dfrac{a_1+a_n}{2}\cdot n\\\\a_1=6;\ n=56;\ a_{56}=391\\\\subtitute\\\\S_{56}=\dfrac{6+391}{2}\cdot56=\dfrac{397}{2}\cdot56=397\cdot28=11,116$

Answer 2

Answer:

The sum of 56 term is 11,116

B is correct.

Step-by-step explanation:

Given:

First term , a=6

Last term l=391

n=56

Formula: Sum of nth term of Arithmetic series.

$Sum=\dfrac{n}{2}(a+l)$

Substitute the value of a, l and n into formula.

$Sum=\dfrac{56}{2}(6+391)$

Sum=11,116

Hence, The sum of 56 term is 11,116