The 10th and 18th terms of an A.P. are 41 and 73 respectively. Find 26th term.

Given:

The 10th and 18th terms of an A.P. are 41 and 73 respectively.

To do:

We have to find the 26th 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,

$a_{10}=a+(10-1)d$

$41=a+9d$

$a=41-9d$......(i)

$a_{18}=a+(18-1)d$

$73=a+17d$

$73=(41-9d)+17d$        (From (i))

$73=41+8d$

$8d=73-41$

$8d=32$

$d=\frac{32}{8}$

$d=4$

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

$a=41-9(4)$

$a=41-36$

$a=5$

26th term of the A.P. $a_{26}=5+(26-1)(4)$

$=5+25(4)$

$=5+100$

$=105$

Hence, the 26th term of the given A.P. is $105$. 

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Updated on: 10-Oct-2022

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