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Q.
The $6^{th}$ term of an $A.P$. is $18$ and the $9^{th}$ term is $12$. The $15^{th}$ term is equal to
Sequences and Series
Solution:
Let $a$ be the first term and $d$ be the common difference.
$\therefore a_6 = 18$
$\Rightarrow a + 5d = 18 \quad....(i)$
$a_9 = 12$
$\Rightarrow a + 8d = 12 \quad...(ii)$
Solving $(i)$ and $(ii)$, we get
$a = 28$,
$ d = - 2$
$\therefore a_{15} = a + 14d$
$ = 28 + 14 (-2)$
$ = 0$