Answer to Question #317744 in Calculus for Fish

Question

Given that the sum of the first four term of an AP is 32, while the sum of the three next terms is 52. Calculate the sum of the first 10 terms of the sequence.

Answer 1

Let $a_1$ be the first term and $d$ be the common difference, i.e. nth term equals

$a_n=a_1+(n-1)d$

The general formula for sum of first n terms is

$S_n=\frac{n}{2}(2a_1+(n-1)d)$

By this formula, the sum of the first 4 terms is

$32=S_4=2(2a_1+3d)=4a_1+6d$

and the sum of the first 7 terms is

$84=32+52=S_7=\frac{7}{2}(2a_1+6d)=7a_1+21d$

By solving two simultaneous equations we can obtain $a_1$ and $d$ :

$\begin{matrix} 4a_1+6d=32 & \Rarr a_1=8-\frac{3}{2}d \\ 7a_1+21d=84 & \Rarr a_1=12-3d \end{matrix}$

Hence

$8-\frac{3}{2}d=12-3d \Rarr (3-\frac{3}{2})d=12-8 \Rarr \frac{3}{2}d=4 \Rarr d=\frac{8}{3}$

Substituting this back into one of equations we can obtain $a_1$ :

$a_1=12-3d=12-3 \times \frac{8}{3}=12-8=4$

Now, having $a_1$ and $d$ , we can compute the sum of the first 10 terms:

$S_{10}=\frac{10}{2}(2a_1+(10-1)d)=5(2\times 4+9\times\frac{8}{3})=160$

ANSWER: 160