The 17th term of an A.P. is 5 more than twice its 8th term. If the 11th term of the A.P. is 43, find the nth term.

Given:

The 17th term of an A.P. is 5 more than twice its 8th term. The 11th term of the A.P. is 43.

To do:

We have to find the nth term.

Solution:

Let the first term of the A.P. be $a$ and the common difference be $d$.

We know that,

nth term of an A.P. $a_n=a+(n-1)d$

Therefore,

$a_{8}=a+(8-1)d$

$=a+7d$......(i)

$a_{17}=a+(17-1)d$

$=a+16d$

According to the question,

$a_{17}=2(a_8)+5$

$a+16d=2(a+7d)+5$

$16d+a=2a+14d+5$

$2a-a=16d-14d-5$

$a=2d-5$.....(i)

$a_{11}=a+(11-1)d$

$43=a+10d$

$43=2d-5+10d$    (From (i))

$12d=43+5$

$d=\frac{48}{12}$

$d=4$

Substituting $d=4$ in (i), we get,

$a=2(4)-5$

$a=8-5$

$a=3$

Therefore,

nth term $a_n=5+(n-1)(4)$

$=3+4n-4$

$=4n-1$

Hence, the nth term of the given A.P. is $4n-1$.  

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Updated on: 10-Oct-2022

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