Which term of the A.P. $3, 15, 27, 39, ….$ will be 120 more than its 21st term?

Given:

Given A.P. is $3, 15, 27, 39, ….$

To do:

We have to find which term of the given A.P. will be 120 more than its 21st term.

Solution:

Here,

$a_1=3, a_2=15, a_3=27$

Common difference $d=a_2-a_1=15-3=12$

We know that,

nth term $a_n=a+(n-1)d$

Therefore,

$a_{21}=3+(21-1)(12)$

$=3+240$

$=243$

120 more than the 21st term $=120+243=363$

This implies,

$a_{n}=3+(n-1)12$

$363=3+12n-12$

$12n=363+9$

$n=\frac{372}{12}$

$n=31$

Hence, 31st term is 120 more than the 21st term. 

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Updated on: 10-Oct-2022

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