The 6th and 17th terms of an A.P. are 19 and 41 respectively, find the 40th term.

Given:

The 6th and 17th terms of an A.P. are 19 and 41 respectively.

To do:

We have to find the 40th 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,

$a_{6}=a+(6-1)d$

$19=a+5d$

$a=19-5d$......(i)

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

$41=a+16d$

$41=(19-5d)+16d$        (From (i))

$41=19+11d$

$11d=41-19$

$11d=22$

$d=\frac{22}{11}$

$d=2$

Substituting $d=2$ in (i), we get,

$a=19-5(2)$

$a=19-10$

$a=9$

40th term of the A.P. $a_{40}=9+(40-1)(2)$

$=9+39(2)$

$=9+78$

$=87$

Hence, the 40th term of the given A.P. is $87$.

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Updated on: 10-Oct-2022

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