If $x=\sum\limits_{n=0}^{\infty }{{{a}^{n}}}$, $y=\sum\limits_{n=0}^{\infty }{{{b}^{n}}}$, $z=\sum\limits_{n=0}^{\infty }{{{c}^{n}}}$ where a, b, c are in A.P and $\left| a \right|<1$, $\left| b \right|<1$, $\left| c \right|<1$, then x, y, z are in:
(a) HP
(b) Arithmetic – Geometric Progression
(c) AP
(d) GP
Answer
299.1k+ views
Hint: We start solving the problem by expanding the given summations of x, y and z. After expanding the summation, we can see that the sum resembles the sum of infinite geometric series. So, we use the sum of the infinite geometric series $\dfrac{p}{1-r}$ if $\left| r \right|<1$ to get the a, b and c in terms of x, y and z. We then use the information given about a, b and c in order to get the desired relation between x, y and z.
Complete step-by-step answer:
According to the problem, we are given $x=\sum\limits_{n=0}^{\infty }{{{a}^{n}}}$, $y=\sum\limits_{n=0}^{\infty }{{{b}^{n}}}$, $z=\sum\limits_{n=0}^{\infty }{{{c}^{n}}}$ where a, b, c are in A.P and $\left| a \right|<1$, $\left| b \right|<1$, $\left| c \right|<1$. We need to find the progression that the numbers x, y, z are in.
Let us first solve for the values of x, y and z.
We have $x=\sum\limits_{n=0}^{\infty }{{{a}^{n}}}$.
$\Rightarrow x={{a}^{0}}+{{a}^{1}}+{{a}^{2}}+{{a}^{3}}+..............\infty $.
$\Rightarrow x=1+a+{{a}^{2}}+{{a}^{3}}+..............\infty $ ---(1).
We can see that the right-hand side resembles the sum of infinite geometric series with the first term as 1 and the common ratio as ‘a’. According the problem we have $\left| a \right|<1$.
We know that the sum of the terms in infinite geometric series $p+pr+p{{r}^{2}}+p{{r}^{3}}+.......\infty $ is $\dfrac{p}{1-r}$ if $\left| r \right|<1$. We use this result in equation (1).
$\Rightarrow x=\dfrac{1}{1-a}$.
$\Rightarrow 1-a=\dfrac{1}{x}$ ---(2).
Now, we have $y=\sum\limits_{n=0}^{\infty }{{{b}^{n}}}$.
$\Rightarrow y={{b}^{0}}+{{b}^{1}}+{{b}^{2}}+{{b}^{3}}+..............\infty $.
$\Rightarrow y=1+b+{{b}^{2}}+{{b}^{3}}+..............\infty $ ---(3).
We can see that the right-hand side resembles the sum of infinite geometric series with first term as 1 and the common ratio as ‘b’. According the problem we have $\left| b \right|<1$.
We know that the sum of the terms in infinite geometric series $p+pr+p{{r}^{2}}+p{{r}^{3}}+.......\infty $ is $\dfrac{p}{1-r}$ if $\left| r \right|<1$. We use this result in equation (3).
$\Rightarrow y=\dfrac{1}{1-b}$.
$\Rightarrow 1-b=\dfrac{1}{y}$ ---(4).
Now, we have $z=\sum\limits_{n=0}^{\infty }{{{c}^{n}}}$.
$\Rightarrow z={{c}^{0}}+{{c}^{1}}+{{c}^{2}}+{{c}^{3}}+..............\infty $.
$\Rightarrow z=1+c+{{c}^{2}}+{{c}^{3}}+..............\infty $ ---(5).
We can see that the right-hand side resembles the sum of infinite geometric series with the first term as 1 and the common ratio as ‘c’. According the problem we have$\left| c \right|<1$.
We know that the sum of the terms in infinite geometric series $p+pr+p{{r}^{2}}+p{{r}^{3}}+.......\infty $ is $\dfrac{p}{1-r}$ if $\left| r \right|<1$. We use this result in equation (5).
$\Rightarrow z=\dfrac{1}{1-c}$.
$\Rightarrow 1-c=\dfrac{1}{z}$ ---(6).
According to the problem, we have a, b and c are in A.P (Arithmetic Progression). We know that if the given terms are in A.P (Arithmetic Progression) and all the terms are multiplied or divided or added or subtracted with any number, then the new terms we obtained will also be in A.P(Arithmetic Progression). Using this fact, we get the numbers –a, –b and –c are in A.P. Also, the numbers $1-a,1-b,1-c$ are also in A.P.
So, we have got 1 – a, 1 – b, 1 – c are in A.P which makes $\dfrac{1}{x}$, $\dfrac{1}{y}$, $\dfrac{1}{z}$ are in A.P.
We know that the inverse of $\dfrac{1}{x}$, $\dfrac{1}{y}$, $\dfrac{1}{z}$ are x, y, z.
We know that if p, q and r are in A.P, then $\dfrac{1}{p}$, $\dfrac{1}{q}$ and $\dfrac{1}{r}$ are in H.P (Harmonic Progression). Using this fact, we get that x, y, z are in H.P.
∴ We have found x, y, z are in H.P.
The correct option for the given problem is (a).
Note: We can prove that when the given terms are in A.P (Arithmetic Progression) and all the terms are multiplied or divided or added or subtracted with any number, then the new terms we obtained will also be in A.P(Arithmetic Progression) by taking example of any series that resembles A.P(Arithmetic Progression). We should not confuse the inverse of A.P as G.P. Similarly, we can expect problems in which x, y and z are in G.P.
Complete step-by-step answer:
According to the problem, we are given $x=\sum\limits_{n=0}^{\infty }{{{a}^{n}}}$, $y=\sum\limits_{n=0}^{\infty }{{{b}^{n}}}$, $z=\sum\limits_{n=0}^{\infty }{{{c}^{n}}}$ where a, b, c are in A.P and $\left| a \right|<1$, $\left| b \right|<1$, $\left| c \right|<1$. We need to find the progression that the numbers x, y, z are in.
Let us first solve for the values of x, y and z.
We have $x=\sum\limits_{n=0}^{\infty }{{{a}^{n}}}$.
$\Rightarrow x={{a}^{0}}+{{a}^{1}}+{{a}^{2}}+{{a}^{3}}+..............\infty $.
$\Rightarrow x=1+a+{{a}^{2}}+{{a}^{3}}+..............\infty $ ---(1).
We can see that the right-hand side resembles the sum of infinite geometric series with the first term as 1 and the common ratio as ‘a’. According the problem we have $\left| a \right|<1$.
We know that the sum of the terms in infinite geometric series $p+pr+p{{r}^{2}}+p{{r}^{3}}+.......\infty $ is $\dfrac{p}{1-r}$ if $\left| r \right|<1$. We use this result in equation (1).
$\Rightarrow x=\dfrac{1}{1-a}$.
$\Rightarrow 1-a=\dfrac{1}{x}$ ---(2).
Now, we have $y=\sum\limits_{n=0}^{\infty }{{{b}^{n}}}$.
$\Rightarrow y={{b}^{0}}+{{b}^{1}}+{{b}^{2}}+{{b}^{3}}+..............\infty $.
$\Rightarrow y=1+b+{{b}^{2}}+{{b}^{3}}+..............\infty $ ---(3).
We can see that the right-hand side resembles the sum of infinite geometric series with first term as 1 and the common ratio as ‘b’. According the problem we have $\left| b \right|<1$.
We know that the sum of the terms in infinite geometric series $p+pr+p{{r}^{2}}+p{{r}^{3}}+.......\infty $ is $\dfrac{p}{1-r}$ if $\left| r \right|<1$. We use this result in equation (3).
$\Rightarrow y=\dfrac{1}{1-b}$.
$\Rightarrow 1-b=\dfrac{1}{y}$ ---(4).
Now, we have $z=\sum\limits_{n=0}^{\infty }{{{c}^{n}}}$.
$\Rightarrow z={{c}^{0}}+{{c}^{1}}+{{c}^{2}}+{{c}^{3}}+..............\infty $.
$\Rightarrow z=1+c+{{c}^{2}}+{{c}^{3}}+..............\infty $ ---(5).
We can see that the right-hand side resembles the sum of infinite geometric series with the first term as 1 and the common ratio as ‘c’. According the problem we have$\left| c \right|<1$.
We know that the sum of the terms in infinite geometric series $p+pr+p{{r}^{2}}+p{{r}^{3}}+.......\infty $ is $\dfrac{p}{1-r}$ if $\left| r \right|<1$. We use this result in equation (5).
$\Rightarrow z=\dfrac{1}{1-c}$.
$\Rightarrow 1-c=\dfrac{1}{z}$ ---(6).
According to the problem, we have a, b and c are in A.P (Arithmetic Progression). We know that if the given terms are in A.P (Arithmetic Progression) and all the terms are multiplied or divided or added or subtracted with any number, then the new terms we obtained will also be in A.P(Arithmetic Progression). Using this fact, we get the numbers –a, –b and –c are in A.P. Also, the numbers $1-a,1-b,1-c$ are also in A.P.
So, we have got 1 – a, 1 – b, 1 – c are in A.P which makes $\dfrac{1}{x}$, $\dfrac{1}{y}$, $\dfrac{1}{z}$ are in A.P.
We know that the inverse of $\dfrac{1}{x}$, $\dfrac{1}{y}$, $\dfrac{1}{z}$ are x, y, z.
We know that if p, q and r are in A.P, then $\dfrac{1}{p}$, $\dfrac{1}{q}$ and $\dfrac{1}{r}$ are in H.P (Harmonic Progression). Using this fact, we get that x, y, z are in H.P.
∴ We have found x, y, z are in H.P.
The correct option for the given problem is (a).
Note: We can prove that when the given terms are in A.P (Arithmetic Progression) and all the terms are multiplied or divided or added or subtracted with any number, then the new terms we obtained will also be in A.P(Arithmetic Progression) by taking example of any series that resembles A.P(Arithmetic Progression). We should not confuse the inverse of A.P as G.P. Similarly, we can expect problems in which x, y and z are in G.P.
Recently Updated Pages
If a parabola whose length of latus rectum is 4a touches class 11 maths JEE_Main

Find the cubic polynomial whose zeroes are 3 5 and class 11 maths JEE_Main

During the sale colour pencils were being sold in -class-11-maths-JEE_Main

A man on the top of a vertical observation tower o-class-11-maths-JEE_Main

In a class of 60 students 25 students play cricket class 11 maths JEE_Main

A regular polygon has 20 sides How many triangles can class 11 maths JEE_Main

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Understanding Atomic Structure for Beginners

Electron Gain Enthalpy and Electron Affinity Explained

Derivation of Equation of Trajectory Explained for Students

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

NCERT Solutions For Class 11 Maths Chapter 6 Permutations And Combinations - 2026-27 Free PDF Download (Login Required)

NCERT Solutions For Class 11 Maths Chapter 9 Straight Lines - 2026-27 Free PDF Download (Sign-in Required)

NCERT Solutions For Class 11 Maths Chapter 8 Sequences And Series - 2026-27 Free PDF Download (Login Required)

NCERT Solutions For Class 11 Maths Chapter 4 Complex Numbers And Quadratic Equations - 2026-27 Free PDF Download (Login Required)

What Are Current and Potential Difference in Electricity?

