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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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