Question

If the ratio of the sum of first three terms and the sum of first six terms of a G.P. be \[125:152\], then the common ratio\[r\]is
A. \[\frac{3}{5}\]
B. \[\frac{5}{3}\]
C. \[\frac{2}{3}\]
D. \[\frac{3}{2}\]

Answer 1

Hint:
A geometric progression (GP) sequence is one in which each subsequent phrase is formed by multiplying each preceding term by a fixed number, or "common ratio." This progression is commonly referred to as a geometric series of integers due to its pattern.
Formula use:
\[{S_n} = \frac{{(a)}{(1 - {r^n)}}}{{1 - r}}\]
A geometric progression has the general form:
\[a,ar,a{r^2},a{r^3},......a{r^n}\]
Where: ‘a’ is the first term; ‘r’ is the common ratio and \[a{r^n}\] is the last term.
Complete step-by-step solution
Every word in a geometric progression or sequence is altered by another term by a common ratio.
A geometric progression has the general form:
\[a,ar,a{r^2},a{r^3},......a{r^n}\]
Where: ‘a’ is the first term; ‘r’ is the common ratio and \[a{r^n}\] is the last term.
Now, we have to consider a G.P:
\[a,ar,a{r^2},a{r^3},......a{r^n}\]
First term will be \[a\]
Second term will be \[ar\]
Third term will be \[a{r^2}\]
Nth term will be \[a{r^{n - 1}}\]
Therefore, the common ratio will be
\[ = \frac{{{\rm{\;any term\;}}}}{{{\rm{\;preceding term\;}}}}\]
That is,
\[ = \frac{{{\rm{\;third term\;}}}}{{{\rm{\;second term\;}}}}\]
\[\frac{{a{r^2}}}{{ar}} = r\]
We know that, sum of n term of GP is
\[{S_n} = \frac{{1 - {r^n}}}{{1 - r}}\]
Now, the sum of first three terms of GP will be
\[{S_3} = a\left( {\frac{{{r^3} - 1}}{{r - 1}}} \right)\]
Then, the sum of first six terms will be
\[{S_6} = a\left( {\frac{{{r^6} - 1}}{{r - 1}}} \right)\]
On solving the above obtained two expressions, we get
\[\frac{{{S_3}}}{{{S_6}}} = \frac{{{r^3} - 1}}{{{r^6} - 1}} = \frac{{125}}{{152}}\]
Now, we get
\[\frac{{{r^3} - 1}}{{({r^3} - 1)({r^3} + 1)}} = \frac{{125}}{{152}}\]
Let’s simplify the above, we get
\[\frac{1}{{({r^3} + 1)}} = \frac{{125}}{{152}}\]
Cross multiplying, we get
\[152 = 125({r^3} + 1)\]
Solve for \[{r^3}\], we get
\[{r^3} = \frac{{27}}{{125}}\]
Thus, we get
\[r = \frac{3}{5}\]
Therefore, the common ratio \[r = \frac{3}{5}\]
Hence, the option A is correct.
Note:
Student must be thorough with the formulas for Geometric progression and their general form to solve these types of problems. And also with the conditions requires solving these types of problems. The outcome will flip between positive and negative if the common ratio is negative. Greater than 1: exponential growth in the direction of infinity (positive). If the number is less than one, there will be an exponential increase to infinity (positive and negative).There will be an exponential decrease towards\[0\]between\[1\]and\[ - 1\]. Zero means, the outcome will remain zero.