Question

Find the sum of the following arithmetic progressions:
$50,46,42,.....$ to $10$ terms

Answer 1

Given A.P is

$50,46,42,...\dots ..$

number of terms of A.P is $n=10$

first term of this A.P is $a_1=50$

second term of this A.P is $a_2=46$

common difference

$d=a_2-a_1=46-50=-4$

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

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

$⟹S_{10}=\frac{10}{2}[2\times 50+(10-1)\times -4]$

$⟹S_{10}=5[100+\left(9\right)\times -4)]$

$⟹S_{10}=5[100-36]$

$⟹S_{10}=5[64]$

$⟹S_{10}=320$

hence the sum of $10$ terms of given A.P is $320$