Question

The sum of first two terms of a $G.P.$ is $1$ and every term of this series is twice of its previous term, then the first term will be:
A. $\dfrac{1}{4}$
B. $\dfrac{1}{3}$
C. $\dfrac{2}{3}$
D. $\dfrac{3}{4}$

Answer 1

In this question firstly we will consider an infinite geometric progression then find out the sum of first two terms that is equals to one after that we will find the value of $r$ for every term that is the twice of sum of successive term and then put the value of $r$ in the above equation obtained to check the correct option.

Complete step by step answer:
We know that the succession of numbers formed according to some definite rule is called a sequence, that is, it is a function whose domain is the set of real numbers. Also a progression is a sequence whose members follow a specific rule of pattern.
A progression is a list of numbers or items that exhibit a particular pattern is called progression.
Progression is of two types:

1. Arithmetic Progression
   2) Geometric Progression
   Arithmetic Progression: A finite or infinite sequence or series ${{a}_{1}}+{{a}_{2}}+.......+{{a}_{n}}$ or ${{a}_{1}}+{{a}_{2}}+.......+{{a}_{n}}+....$ is said to be an arithmetic progression $(A.P)$ ${{a}_{k}}-{{a}_{k-1}}=d$, a constant independent of $k$ for $k=1,2,3,4,.......,n$ or $k=2,3,4,.....$ Hence the constant $d$ is called the common difference of the arithmetic progression.
   Geometric Progression: A sequence is called a geometric progression if the ratio of any term to the preceeding terms is a constant called common ratio that is a finite sequence ${{a}_{1}}+{{a}_{2}}+.......+{{a}_{n}}$ or ${{a}_{1}}+{{a}_{2}}+.......+{{a}_{n}}+....$ is said to be geometric progression if none of the ${{a}_{n}}$ is zero and that $\dfrac{{{a}_{k+1}}}{{{a}_{k}}}=r$, a constant $\text{(independent of k)}$ for $k=1,2,3,4,...$
   The first term and the common ratio of $G.P.$ are denoted as $a,r$
   Then ${{n}^{th}}$term of a $G.P.$ is given by the formula:
   ${{a}_{n}}=a{{r}^{n-1}}$
   According to the question:
   Let the given infinite $G.P.$ is $a,ar,a{{r}^{2}},a{{r}^{3}}.....\infty$
   It is given that the sum of the first two terms of an infinite $G.P.$ is $1$.
   Hence, $a+ar=1$
   Now every term is the twice of sum of successive term:
   \Rightarrow $$$$a=2(ar+a{{r}^{2}}+a{{r}^{3}}.......\infty )
   So, $a=2(\dfrac{ar}{1-r})$
   Now cancel out $a$ from both sides we get:
   $\Rightarrow$ $1=2(\dfrac{r}{1-r})$
   $\Rightarrow$ $1-r=2r$
   So, $r=\dfrac{1}{3}$
   Now putting the value of $r$ in the equation $a+ar=1$ to find the first term we will get:
   $\Rightarrow$ $a+a\times \dfrac{1}{3}=1$
   $\Rightarrow$ $\dfrac{3a+a}{3}=1$
   $\Rightarrow$ $\dfrac{4a}{3}=1$
   $\Rightarrow$ $a=\dfrac{3}{4}$

So, the correct answer is “Option D”.

Note: We must remember that if each term of a geometric progression be multiplied by a nonzero number, then the sequence obtained is also a geometric progression and if we multiply the corresponding terms of two geometric progressions then he sequence obtained is also a geometric progression.