Question

If the sum of the second, third and fourth terms of a positive term G.P. is 3 and the sum of its sixth, seventh and eighth terms is 243, then the sum of the first 50 terms of this G.P. is
1) $(2 / 13)\left(3^{50}-1\right)$
2) $(1 / 26)\left(3^{49}-1\right)$
3) $(1 / 13)\left(3^{50}-1\right)$
4) $(1 / 26)\left(3^{50}-1\right)$

Answer 1

Hint: A sort of sequence known as geometric progression (GP) is one in which each following phrase is created by multiplying each preceding term by a fixed number, or "common ratio." This progression is sometimes referred to as a pattern-following geometric sequence of numbers.

Formula Used:
The general expression is $a+ar+a{{r}^{2}}+a{{r}^{3}}$

Complete step by step Solution:
Let first term $=\mathrm{a}>0$
Common ratio $=\mathrm{r}>0$
When one term is varied by another by a common ratio, the series is referred to as a geometric progression or sequence. When we multiply the previous term by a constant (which is non-zero), we get the following term in the sequence. It is symbolised by:
$a+ar+a{{r}^{2}}+a{{r}^{3}}$and so on
where $r$ is the common ratio and $a$ is the first term.
$a r+a r^{2}+a r^{3}=3$………….(1)
$a r^{5}+a r^{6}+a r^{7}=243$…………(2)
$r^{4}\left(a r+a r^{2}+a r^{3}\right)=243$
$r^{4}(3)=243 \Rightarrow r=3$ as $r>0$
from (1)
$3 a+9 a+27 a=3$
$a=\dfrac{1}{13}$
$S_{n}=\dfrac{a\left(r^{n}-1\right)}{(r-1)}$
$S_{50}=\dfrac{a\left(r^{50}-1\right)}{(r-1)}=\dfrac{1}{26}\left(3^{50}-1\right)$

Hence, the correct option is 4.

Note: When a geometric progression (GP) is finite, it may be represented as $a r^{2}, a r^{3}, \ldots a r^{n-1}$ and when it is infinite, it can be expressed as$a r^{2}, a r^{3}, \ldots a r^{n-1}$. Using a few formulas, we can determine the sum to n terms of GP for both finite and infinite GP. Additionally, the equation for calculating the sum of the finite and finite GP separately can be derived. The first n terms of a GP are collectively referred to as the total to n terms of a GP.