Question

The first term of an infinite G.P. is $ 1 $ and every term is equal to the sum of the successive terms, then its fourth term will be:
 $
  A.\,\,\dfrac{1}{2} \\
  B.\,\,\dfrac{1}{8} \\
  C.\,\,\dfrac{1}{4} \\
  D.\,\,\dfrac{1}{{16}} \\
  $

Answer 1

Hint: To find the fourth term of the geometric progression we first let consider an infinite G.P. series in which the first term is $ 1 $ and then equating the first term equal to sum of all successive terms and then finding the sum of infinite G.P. so formed to get the value of ‘r’ and hence different terms of infinite series and hence required solution of the given problem.
Formulas used: Sum of infinite G.P. = $ \dfrac{a}{{1 - r}}, $ where ‘a’ is first term and ‘r’ is the common difference of G.P.

Complete step by step solution:
The first term of G.P. is $ 1 $ .
So, let's consider a G.P. series whose terms are given as:
 $ 1,r,{r^2},{r^3}.............\infty $
Also, it is given that every term of a G.P. series is equal to the sum of the successive terms.
Therefore, we can write the above term of G.P. series as:
 $ r + {r^2} + {r^3} + ............\infty = 1 $
Using the infinite sum formula of G.P. on the left side of the above equation. We have,
 $ \dfrac{r}{{1 - r}} = 1$
$\left\{ {{S_\infty } = \dfrac{a}{{1 - r}}} \right\} $
 $
   \Rightarrow r = 1 - r \\
   \Rightarrow r + r = 1 \\
   \Rightarrow 2r = 1 \\
   \Rightarrow r = \dfrac{1}{2} \\
  $
Therefore, the common ratio of G.P. is $ \dfrac{1}{2}. $
Substituting value of $ r = \dfrac{1}{2} $ in above series. We have,
 $ 1,\dfrac{1}{2},\dfrac{1}{4},\dfrac{1}{8},\dfrac{1}{{16}},.......... $
Hence, fourth term of the above G.P. series is $ \dfrac{1}{8} $ .
So, the correct answer is “Option B”.

Note: Geometric progression is a series in which two consecutive terms are multiple of some number and there are different formulas to find sum of terms in G.P. as if there are finite numbers of terms or we can say if there are ‘n’ terms given then formula of sum is given as: $ {S_n} = a\left( {\dfrac{{{r^n} - 1}}{{r - 1}}} \right),r > 1\,\,or\,\,{S_n}a\left( {\dfrac{{1 - {r^n}}}{{1 - r}}} \right),\,\,r < 1 $ and for infinite G.P. formula for sum of terms is $ {S_\infty } = \dfrac{a}{{1 - r}} $ . So, one should choose the correct formula of sum to find the required solution to a given problem.