Question

Which term of the A.P. $3,10,17,......$ wil be $84$ more than its $13$th term?

Answer 1

Given A.P

$3,10,17,...\dots ..$

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

second term of A.p is $a_2=10$

common difference $d=a_2-a_1=10-3=7$

nth term of A.P is given by

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

$a_n=3+(n-1)7$

$a_n=3+7n-7$

$a_n=-4+7n...eq\left(1\right)$

13th term of A.P is

$a_{13}=3+(13-1)7$

$a_{13}=3+\left(12\right)7$

$a_{13}=3+84$

$a_{13}=\mathrm{87....}eq\left(2\right)$

we have to find term which is 84 more than 13th term

$⟹a_n=a_{13}+84$

put value of $a_n$ from eq(1) and $a_{13}=87$ from eq(2) in above equation

$⟹-4+7n=87+84$

$⟹7n=87+84+4$

$⟹7n=175$

$⟹n=\frac{175}{7}$

$⟹n=25$

hence $25$th term of this A.P will be $84$ more than its $13$th term