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

Question:

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

Solution:

In the given problem, let us first find the 13th term of the given A.P.

A.P. is 3, 10, 17 …

Here,

First term (a) = 3

Common difference of the A.P. $(d)=10-3=7$

Now, as we know,

$a_{n}=a+(n-1) d$

So, for 13th term (n = 13),

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

$=3+12(7)$

$=3+84$

 

$=87$

Let us take the term which is 84 more than the 13th term as an. So,

$a_{n}=84+a_{13}$

$=84+87$

 

$=171$

Also, $a_{n}=a+(n-1) d$

$171=3+(n-1) 7$

 

$171=3+7 n-7$

$171=-4+7 n$

$171+4=7 n$

Further simplifying, we get,

$175=7 n$

$n=\frac{175}{7}$

$n=25$

Therefore, the $25^{\text {th }}$ term of the given A.P. is 84 more than the $13^{\text {th }}$ term.

 

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