Question

Find the sum of the following $APS$(a)$2,7,12,$ to $10$ terms(b)$\frac{1}{15},\frac{1}{12},\frac{1}{10}$ to $11$ terms.

Answer 1

$\left(a\right)$

$2,7,12\dots$

$n=10,a=2,d=5$

$S_{10}=\frac{10}{2}[2a+9d]=5[4+45]=245$

$\left(b\right)$

$\frac{1}{15},\frac{1}{12},\frac{1}{10}\dots$

$a=\frac{1}{15},d=\frac{1}{60}$

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

$S_{11}=\frac{11}{2}[2\times \frac{1}{15}+10\times \frac{1}{60}]$

$S_{11}=\frac{11}{2}[\frac{2}{15}+\frac{1}{6}]$

$S_{11}=\frac{11}{2}\times \frac{3}{10}=\frac{33}{20}$