Question

Find the sum of the arithmetic progressions: $3,\frac{9}{2},16,\frac{15}{2},......$ to $25$ terms

Answer 1

Given A.P is

$3,\frac{9}{2},16,\frac{15}{2}..\dots ..$

number of terms of A.P is $n=25$

first term of this A.P is $a_1=3$

second term of this A.P is $a_2=\frac{9}{2}$

common difference

$d=a_2-a_1=\frac{9}{2}-3=\frac{3}{2}$

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

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

$⟹S_{25}=\frac{25}{2}[2\times 3+(25-1)\times \frac{3}{2}]$

$⟹S_{25}=12.5[6+\left(24\right)\times \frac{3}{2}]$

$⟹S_{25}=12.5[6+36]$

$⟹S_{25}=12.5[42]$

$⟹S_{25}=525$

hence the sum of $25$ terms of given A.P is $525$