Question

# Find the 15th term of the G.P. $3,12,48,192,.....$a. $3 \times {4^{15}}$b. $3 \times {4^{14}}$c. $3 \times {4^{16}}$d. ${3^{15}}$

Hint: In this problem to get our desired result, we need to find the 1st term and also find the common ratio of the given G.P. series. Then using the formula of the general term of G.P., we will get our solution.

Complete step by step answer:

Here, in the given problem, we are to find the 15th term of the G.P. $3,12,48,192,.....$
So, we have our first term as,$a = 3$, then the common ratio is obtained by second term divided by first term, i.e. $\dfrac{{12}}{3}$$= 4$
As per the formula of the G.P,
${T_n} = a{r^{n - 1}}$
The 15th term would be, $a{r^{15 - 1}}$$= a{r^{14}}$, where a is the first term and the common ratio r.
Now, we have, $a = 3,r = 4$, the 15th term would be, $3 \times {4^{14}}$.

Note: In mathematics, a geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. For example, sequence 2, 6, 18, 54, ... is a geometric progression with common ratio 3.