Question

If b is the first term of an infinite G.P, whose sum is five, then b lies in the interval.
\[
  (a){\text{ (}} - \infty , - 10] \\
  (b){\text{ }}\left( {10,\infty } \right) \\
  (c){\text{ }}\left( {0,10} \right) \\
  (d){\text{ }}\left( { - 10,0} \right) \\
 \]

Answer 1

Hint – In this question use the direct formula for the sum of infinite numbers of a G.P which is ${S_\infty } = \dfrac{a}{{1 - r}}$, along with the concept that the common ratio r always lies between $ - 1 < r < 1$, to get the desired range of b.

Complete Step-by-Step solution:
As we know for an infinite G.P the sum (${S_\infty }$) of infinite terms is given as
${S_\infty } = \dfrac{a}{{1 - r}}$............................ (1)
Where, a = first term of an G.P
               r = common ratio of G.P.
And we know the common ratio for an infinite G.P always lies between (-1 to 1).
$ \Rightarrow - 1 < r < 1$
Now subtract by (-1) in above equation we have
$ \Rightarrow - 1 - 1 < r - 1 < 1 - 1$
$ \Rightarrow - 2 < r - 1 < 0$
Now multiply by (-1) when we multiply the inequality sign reversed so we have,
$ \Rightarrow 2 > 1 - r > 0$............................. (2)
Now it is given that the first term of an G.P is (b) and the sum is 5 therefore from equation (1) we have,
$ \Rightarrow 5 = \dfrac{b}{{1 - r}}$
$ \Rightarrow \left( {1 - r} \right) = \dfrac{b}{5}$
Now from equation (2) we have,
$ \Rightarrow 2 > \dfrac{b}{5} > 0$
Now multiply by 5 we have
$ \Rightarrow 10 > b > 0$
$ \Rightarrow 0 < b < 10$
Therefore b lies in the interval (0 to 10)
So this is the required answer.
Hence option (C) is correct.
Note – A series is said to be in G.P (geometric progression) if and only if the common ratio that is the ratio between the two consecutive terms of that series remains constant throughout the series. It is advised to remember the direct formula for series like, arithmetic progression, harmonic progression and geometric progression as these are the most frequently used series.