Question

Find the term of the arithmetic progression $9,12,15,18,.....$ which is $39$ more than its $36$th term.

Answer 1

Given A.P

$9,12,15,18,...\dots ..$

first term of A.P is $a_1=9$

second term of A.p is $a_2=12$

common difference $d=a_2-a_1=12-9=3$

nth term of A.P is given by

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

$a_n=9+(n-1)3$

$a_n=9+3n-3$

$a_n=6+3n...eq\left(1\right)$

36th term of A.P is given by putting n=36 in eq(1)

$a_{36}=6+3\times 36$

$a_{36}=6+108$

$a_{36}=\mathrm{114....}eq\left(2\right)$

we have to find term which is 39 more than 36th term

$⟹a_n=a_{36}+39$

put value of $a_n$ from eq(1) and $a_{36}=114$ from eq(2) in above equation

$⟹6+3n=114+39$

$⟹3n=114+39-6$

$⟹3n=147$

$⟹n=\frac{147}{3}$

$⟹n=49$

hence $49$th term of this A.P will be $39$ more than its $36$th term