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Question: In a geometric progression the ratio of the sum of the first three terms and the first six terms is ...

In a geometric progression the ratio of the sum of the first three terms and the first six terms is 125:152125:152. The common ratio
A.15\dfrac{1}{5}
B.25\dfrac{2}{5}
C.45\dfrac{4}{5}
D.35\dfrac{3}{5}

Explanation

Solution

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,ar2,ar3,...,arna,ar,a{r^2},a{r^3},...,a{r^n}
where aa is the first term
rr is the common ratio
arna{r^n}is the last term
Consider a GP a,ar,ar2,ar3,...,arna,ar,a{r^2},a{r^3},...,a{r^n}
First term =a = a
Second term =ar = ar
Third term =ar2 = a{r^2}
Nth term =arn−1 = a{r^{n - 1}}
Therefore common ratio =any termpreceding term = \dfrac{{any\,term}}{{preceding\,term}}
=third termsec⁡ond term= \dfrac{{third\,term}}{{\sec ond\,term}}
=ar2ar=r= \dfrac{{a{r^2}}}{{ar}} = r
Sum of nnterms of a GP Sn=1−rn1−r{S_n} = \dfrac{{1 - {r^n}}}{{1 - r}}
If the common ratio is:
Negative: the result will alternate between positive and negative.
Greater than 11 : there will be an exponential growth towards infinity (positive).
Less than −1 - 1 : there will be an exponential growth towards infinity (positive and negative).
Between 11 and −1 - 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, S3=a(r3−1r−1){S_3} = a\left( {\dfrac{{{r^3} - 1}}{{r - 1}}} \right)
The sum of the first six terms of GP, S6=a(r6−1r−1){S_6} = a\left( {\dfrac{{{r^6} - 1}}{{r - 1}}} \right)
Now
S3S6=r3−1r6−1=125152\dfrac{{{S_3}}}{{{S_6}}} = \dfrac{{{r^3} - 1}}{{{r^6} - 1}} = \dfrac{{125}}{{152}}
Therefore we get
r3−1(r3−1)(r3+1)=125152\dfrac{{{r^3} - 1}}{{\left( {{r^3} - 1} \right)\left( {{r^3} + 1} \right)}} = \dfrac{{125}}{{152}}
On simplifying we get
1(r3+1)=125152\dfrac{1}{{\left( {{r^3} + 1} \right)}} = \dfrac{{125}}{{152}}
On cross multiplication we get
152=125(r3+1)152 = 125\left( {{r^3} + 1} \right)
Hence on solving we get
r3=27125{r^3} = \dfrac{{27}}{{125}}
Hence we get
r=35r = \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.