Question:
If 7th and 13th terms of an A.P. be 34 and 64 respectively, then its 18th term is
(a) 87
(b) 88
(c) 89
(d) 90
Solution:
(c) 89
$a_{7}=34$
$\Rightarrow a+6 d=34$ ....(1)
Also, $a_{13}=64$
$\Rightarrow a+12 d=64$ ....(2)
Solving equations $(1)$ and $(2)$, we get:
a = 4 and d = 5
$\therefore a_{18}=a+17 d$
$=4+17(5)$
$=89$
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