The 19th term of an AP is equal to 3 times its 6th term. If its 9th term is 19, find the AP.

Solution:

Let a be the first term and d be the common difference of the AP. Then,

$a_{19}=3 a_{6}$             (Given)

$\Rightarrow a+18 d=3(a+5 d) \quad\left[a_{n}=a+(n-1) d\right]$

$\Rightarrow a+18 d=3 a+15 d$

$\Rightarrow 3 a-a=18 d-15 d$

$\Rightarrow 2 a=3 d$                 $\ldots \ldots(1)$

Also,

$a_{9}=19$             (Given)

$\Rightarrow a+8 d=19 \quad \ldots \ldots(2)$

From (1) and (2), we get

$\frac{3 d}{2}+8 d=19$

$\Rightarrow \frac{3 d+16 d}{2}=19$

$\Rightarrow 19 d=38$

$\Rightarrow d=2$

Putting d = 2 in (1), we get

$2 a=3 \times 2=6$

$\Rightarrow a=3$

So,

$a_{2}=a+d=3+2=5$

$a_{3}=a+2 d=3+2 \times 2=7, \ldots$

Hence, the AP is 3, 5, 7, 9, ... .