Question

Find the sum of the following arithmetic progressions:
$41,36,31,.....$ to $12$ terms.

Answer 1

Given A.P is

$41,36,31,...\dots ..$

number of terms of A.P is $n=12$

first term of this A.P is $a_1=41$

second term of this A.P is $a_2=36$

common difference

$d=a_2-a_1=36-41=-5$

we know that sum of n term of an A.P is given by

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

$⟹S_{12}=\frac{12}{2}[2\times 41+(12-1)\times -5]$

$⟹S_{12}=6[82+\left(11\right)\times -5)]$

$⟹S_{12}=6[82-55]$

$⟹S_{12}=6[27]$

$⟹S_{12}=162$

hence the sum of 12 terms of given A.P is $162.$