Question

Find the $12$th term from the end of the following arithmetic progressions.
$1,4,7,10,..........,88$

Answer 1

Given A.P

$1,4,7,10,...\dots ..,88$

first term of this A.P $a_1=1$

second term of this A.P $a_2=4$

common difference $d=a_2-a_1=4-1=3$

the nth term of an A.P is given by

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

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

$a_n=1+3n-3$

$a_n=-2+3n\dots ..eq\left(1\right)$

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

put $a_n=88$ the last term of A.P in above eq(1)

$88=-2+3n$

$3n=88+2$

$n=\frac{90}{3}$

$⟹n=30$

hence this A.P contains $30$ terms

for calculating $12$th term from the end means $(30-11)$th term from starting of A.P

hence we have to calculate 19th term of this A.P

so put $n=19$ in eq(1)

$a_{19}=-2+3\times 19$

$a_{19}=-2+57$

$a_{19}=55$

hence $12$th term of from the end of this A.P is $55$