Question

In a geometric progression the ratio of the sum of the first three terms and the first six terms is \[125:152\]. The common ratio
A.\[\dfrac{1}{5}\]
B.\[\dfrac{2}{5}\]
C.\[\dfrac{4}{5}\]
D.\[\dfrac{3}{5}\]

Answer 1

Hint: Geometric Progression (GP) is a type of sequence where each succeeding term is produced by multiplying each preceding term by a fixed number, which is called a common ratio. This progression is also known as a geometric sequence of numbers as it follows a pattern.

Complete answer:
A geometric progression or a geometric sequence is the one, in which each term is varied by another by a common ratio. General form of a GP is
\[a,ar,a{r^2},a{r^3},...,a{r^n}\]
where \[a\] is the first term
\[r\] is the common ratio
\[a{r^n}\]is the last term
Consider a GP \[a,ar,a{r^2},a{r^3},...,a{r^n}\]
First term \[ = a\]
Second term \[ = ar\]
Third term \[ = a{r^2}\]
Nth term \[ = a{r^{n - 1}}\]
Therefore common ratio \[ = \dfrac{{any\,term}}{{preceding\,term}}\]
\[ = \dfrac{{third\,term}}{{\sec ond\,term}}\]
\[ = \dfrac{{a{r^2}}}{{ar}} = r\]
Sum of \[n\]terms of a GP \[{S_n} = \dfrac{{1 - {r^n}}}{{1 - r}}\]
If the common ratio is:
Negative: the result will alternate between positive and negative.
Greater than \[1\] : there will be an exponential growth towards infinity (positive).
Less than \[ - 1\] : there will be an exponential growth towards infinity (positive and negative).
Between \[1\] and \[ - 1\]: there will be an exponential decay towards zero.
Zero: the result will remain at zero
The sum of the first three terms of GP, \[{S_3} = a\left( {\dfrac{{{r^3} - 1}}{{r - 1}}} \right)\]
The sum of the first six terms of GP, \[{S_6} = a\left( {\dfrac{{{r^6} - 1}}{{r - 1}}} \right)\]
Now
\[\dfrac{{{S_3}}}{{{S_6}}} = \dfrac{{{r^3} - 1}}{{{r^6} - 1}} = \dfrac{{125}}{{152}}\]
Therefore we get
\[\dfrac{{{r^3} - 1}}{{\left( {{r^3} - 1} \right)\left( {{r^3} + 1} \right)}} = \dfrac{{125}}{{152}}\]
On simplifying we get
\[\dfrac{1}{{\left( {{r^3} + 1} \right)}} = \dfrac{{125}}{{152}}\]
On cross multiplication we get
\[152 = 125\left( {{r^3} + 1} \right)\]
Hence on solving we get
\[{r^3} = \dfrac{{27}}{{125}}\]
Hence we get
\[r = \dfrac{3}{5}\]
Therefore, option (D) is the correct answer.

Note:
Geometric Progression (GP) is a type of sequence where each succeeding term is produced by multiplying each preceding term by a fixed number, which is called a common ratio. This progression is also known as a geometric sequence of numbers as it follows a pattern.