Question

If the sum of a certain number of terms starting from first term of an A.P is $25,22,19,.....$ is $116$. Find the last term.

Answer 1

Given A.P is

$25,22,19,...\dots ..$

sum of $n$ terms of A.P is $S_n=116$

first term of this A.P is $a_1=25$

second term of this A.P is $a_2=22$

common difference

$d=a_2-a_1=22-25=-3$

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

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

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

$⟹S_n=\frac{n}{2}[50-3n+3)]$.

$⟹S_n=\frac{n}{2}[53-3n]$

put value of $S_n=116$ in above equation

$⟹116=\frac{n}{2}[53-3n]$

$⟹[53n-3n^2]=232$

$3n^2-53n+232=0$

by solving the above quadratic equation we get

$n=(53\pm \sqrt{(-53{)}^2-4\times 3\times 232})/6$

$n=(53\pm 5)/6$

$n=\frac{58}{6}=9.6666$ which is not possible since n must be an integer.

or

$n=\frac{48}{6}=8$

hence the sum of 8 term of this A.P is 116.

we know that sum of n term is also given by

$S_n=\frac{n}{2}(a+l)$ where l is the last term of A.P

put value of $S_n=116;a=25;n=8$ we get

$116=\frac{8}{2}(25+l)$

$25+l=29$

$⟹l=29-25=4$

hence last term of A.P is $4$