Question

If S is the sum of an infinite G.P , the first term \[a\] then the common ratio \[r\] given by
\[\left( 1 \right)\] \[\dfrac{{a - s}}{s}\]
\[\left( 2 \right)\] \[\dfrac{{s - a}}{s}\]
\[\left( 3 \right)\] \[\dfrac{a}{{1 - s}}\]
\[\left( 4 \right)\] \[\dfrac{{s - a}}{a}\]

Answer 1

Hint: We have to find the common ratio \[r\] of a G.P. . We solve this question using the concept of sum of infinite series of geometric progression (G.P.) . From the given values of the first term (\[a\]) we can compute the common ratio (\[r\]) . And using the formula of infinite series of G.P. we can find the common ratio .

Complete step-by-step solution:
Given :
\[s\] is the sum of an infinite G.P.
First term of the G.P. is \[a\]
We know , that the formula of sum of infinite series is given as :
\[{S_\infty } = \dfrac{a}{{1 - r}}\]
Where \[\left| r \right| < 1\] .
Putting the values in the formula of sum of infinite series and simplifying the expression , we get
\[s = \dfrac{a}{{1 - r}}\]
Cross multiplying the terms , we can write the expression as :
\[s\left( {1 - r} \right) = a\]
\[s - sr = a\]
Also we can write the expression as :
\[s + a = sr\]
Also , we can find the relation as :
\[r = \dfrac{{s - a}}{s}\]
Thus , the value of common ratio (\[r\]) of the infinite G.P. is \[\dfrac{{s - a}}{s}\] .
Hence , the correct option is \[\left( 2 \right)\].

Note: For the terms of a given series to be in G.P. the common ratio between the terms of the series should be the same for all the two consecutive terms of the series . The ratio of the second term to the first term of the given series should be the same as that of the ratio of the third term to the second term of the given series .
The sum of \[n\] terms of a G.P. is given by the formula :
\[{S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}\]
The expression for \[{n^{th}}\] term of G.P. is given by the formula :
\[{a_n} = a \times {r^{n - 1}}\]