Question

Find the $12$th term from the end of the following arithmetic progressions.
$3,8,13,......,253$

Answer 1

Given A.P

$3,8,13,...\dots ..,253$

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

second term of this A.P $a_2=8$

common difference $d=a_2-a_1=8-3=5$

the nth term of an A.P is given by

$a_n=a_1+(n-1)d$

$a_n=3+(n-1)5$

$a_n=3+5n-5$

$a_n=-2+5n\dots ..eq\left(1\right)$

first calculate number of terms(n) in this A.P

put $a_n=253$ the last term of A.P in above eq(1)

$253=-2+5n$

$5n=253+2$

$n=\frac{255}{5}$

$⟹n=51$

hence this A.P contains $51$terms

for calculating $12$th term from the end means $(51-11)$th term from starting of A.P

hence we have to calculate 40th term of this A.P

so put $n=40$ in eq(1)

$a_{40}=-2+5\times 40$

$a_{40}=-2+200$

$a_{40}=198$

hence $12$th term of from the end of this A.P is $198$