
If \[x = \sum\limits_{n = 0}^{n = \infty } {{a^n}} ,y = \sum\limits_{n = 0}^{n = \infty } {{b^n}} ,z = \sum\limits_{n = 0}^{n = \infty } {{c^n}} \]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) GP
B) AP
C) HP
D) None of these
Answer
162.6k+ views
Hint: in this question we have to find type series follow by the given terms. Use sum infinite term of GP to find value of x, y, z. Use concept of AP according to which twice the middle term equal to sum of first term and last term.
Formula Used: Sum of infinite term of GP calculated by using
\[{S_\infty } = \dfrac{a}{{1 - r}}\]
Where
a is first term of GP
r is common ratio of GP
Complete step by step solution: Given: \[x = \sum\limits_{n = 0}^{n = \infty } {{a^n}} ,y = \sum\limits_{n = 0}^{n = \infty } {{b^n}} ,z = \sum\limits_{n = 0}^{n = \infty } {{c^n}} \]where a, b, c are in A.P and \[\left| a \right| < 1,\left| b \right| < 1,\left| c \right| < 1\;\]
\[x = \sum\limits_{n = 0}^{n = \infty } {{a^n} = {S_\infty } = \dfrac{a}{{1 - r}}} \]
Here common ratio is equal to a
\[x = {S_\infty } = \dfrac{1}{{1 - a}}\]
\[\dfrac{1}{x} = 1 - a\]
\[y = \sum\limits_{n = 0}^{n = \infty } {{b^n} = {S_\infty } = \dfrac{a}{{1 - r}}} \]
Here common ratio is equal to b
\[y = {S_\infty } = \dfrac{1}{{1 - b}}\]
\[\dfrac{1}{y} = 1 - b\]
\[z = \sum\limits_{n = 0}^{n = \infty } {{c^n} = {S_\infty } = \dfrac{a}{{1 - r}}} \]
Here common ratio is equal to c
\[z = {S_\infty } = \dfrac{1}{{1 - c}}\]
\[\dfrac{1}{z} = 1 - c\]
\[\dfrac{1}{y} - \dfrac{1}{x} = a - b\] And \[\dfrac{1}{z} - \dfrac{1}{y} = b - c\]
It is given in the question that a, b, c are in AP therefore \[a - b = b - c\]
\[\dfrac{1}{y} - \dfrac{1}{x} = \dfrac{1}{z} - \dfrac{1}{y}\]
\[\dfrac{2}{y} = \dfrac{1}{x} + \dfrac{1}{z}\]
So, we can say that x, y and z are in HP.
Option ‘C’ is correct
Note: Here we must remember that if twice the reciprocal of middle term equal sum of reciprocal of first term and third term then three terms are in HP. If twice the middle term equal to sum of first term and last term then three terms are in AP.
Sometime students get confused in between AP and GP the only difference in between them is in AP we talk about common difference whereas in GP we talk about common ratio.
Formula Used: Sum of infinite term of GP calculated by using
\[{S_\infty } = \dfrac{a}{{1 - r}}\]
Where
a is first term of GP
r is common ratio of GP
Complete step by step solution: Given: \[x = \sum\limits_{n = 0}^{n = \infty } {{a^n}} ,y = \sum\limits_{n = 0}^{n = \infty } {{b^n}} ,z = \sum\limits_{n = 0}^{n = \infty } {{c^n}} \]where a, b, c are in A.P and \[\left| a \right| < 1,\left| b \right| < 1,\left| c \right| < 1\;\]
\[x = \sum\limits_{n = 0}^{n = \infty } {{a^n} = {S_\infty } = \dfrac{a}{{1 - r}}} \]
Here common ratio is equal to a
\[x = {S_\infty } = \dfrac{1}{{1 - a}}\]
\[\dfrac{1}{x} = 1 - a\]
\[y = \sum\limits_{n = 0}^{n = \infty } {{b^n} = {S_\infty } = \dfrac{a}{{1 - r}}} \]
Here common ratio is equal to b
\[y = {S_\infty } = \dfrac{1}{{1 - b}}\]
\[\dfrac{1}{y} = 1 - b\]
\[z = \sum\limits_{n = 0}^{n = \infty } {{c^n} = {S_\infty } = \dfrac{a}{{1 - r}}} \]
Here common ratio is equal to c
\[z = {S_\infty } = \dfrac{1}{{1 - c}}\]
\[\dfrac{1}{z} = 1 - c\]
\[\dfrac{1}{y} - \dfrac{1}{x} = a - b\] And \[\dfrac{1}{z} - \dfrac{1}{y} = b - c\]
It is given in the question that a, b, c are in AP therefore \[a - b = b - c\]
\[\dfrac{1}{y} - \dfrac{1}{x} = \dfrac{1}{z} - \dfrac{1}{y}\]
\[\dfrac{2}{y} = \dfrac{1}{x} + \dfrac{1}{z}\]
So, we can say that x, y and z are in HP.
Option ‘C’ is correct
Note: Here we must remember that if twice the reciprocal of middle term equal sum of reciprocal of first term and third term then three terms are in HP. If twice the middle term equal to sum of first term and last term then three terms are in AP.
Sometime students get confused in between AP and GP the only difference in between them is in AP we talk about common difference whereas in GP we talk about common ratio.
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