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

Given:

Given A.P. is $3, 10, 17, …$

To do:

We have to find which term of the given A.P. will be 84 more than its 13th term.

Solution:

Here,

$a_1=3, a_2=10, a_3=17$

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

We know that,

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

Therefore,

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

$=3+84$

$=87$

84 more than the 13th term $=84+87=171$

This implies,

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

$171=3+7n-7$

$7n=171+4$

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

$n=25$

Hence, 25th term is 84 more than the 13th term.

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