Question

If first and \[{{\left( 2n-1 \right)}^{th}}\] term of an AP, GP and HP are equal and their \[{{n}^{th}}\] terms are a, b, c respectively, then,
(This question has multiple correct options)
\[\left( \text{a} \right)\text{ }a+c=2b\]
\[\left( \text{b} \right)\text{ }a\ge b\ge c\]
\[\left( \text{c} \right)\text{ }a+c=b\]
\[\left( \text{d} \right)\text{ }{{b}^{2}}=ac\]

Answer 1

Hint: In this question, assume the first term as x and \[{{\left( 2n-1 \right)}^{th}}\] term as y for AP, GP and HP. Now, find the middle term or the mean using these terms which would be \[{{n}^{th}}\] term of each sequence and equate them to a, b and c accordingly to find the relation between them.

Complete step-by-step answer:
In this question, we are given that the first and \[{{\left( 2n-1 \right)}^{th}}\] term of an AP, GP and HP are equal and their \[{{n}^{th}}\] terms are a, b, c respectively. We have to find the relation between a, b and c.
We are given that the first term of AP, GP and HP are equal. So, let us consider the first term of all the sequences as x. We are also given that \[{{\left( 2n-1 \right)}^{th}}\] terms of AP, GP and HP are equal. So, let us consider the \[{{\left( 2n-1 \right)}^{th}}\] term of all the sequences as y.
We know that the middle term of any sequence is known as its mean. We know that if ‘a’ is the first number and b is the last number, then the mean is given by \[\dfrac{a+b}{2}.\] Similarly, the middle term or the mean of first and \[{{\left( 2n-1 \right)}^{th}}\] term is given by,
\[\text{Mean}=\dfrac{1+\left( 2n-1 \right)}{2}=\dfrac{2n}{2}={{n}^{th}}\text{ term}\]
So, we get the mean of the first and \[{{\left( 2n-1 \right)}^{th}}\] term as the \[{{n}^{th}}\] term. Now, we know that the arithmetic mean (AM) of two numbers say a and b is given by \[\dfrac{a+b}{2}.\] So, we get, AM of its first term x and \[{{\left( 2n-1 \right)}^{th}}\] term y is given by \[\dfrac{x+y}{2}.\]
We are also given that the \[{{n}^{th}}\] term of AP is a. So, we get,
\[AM=\dfrac{x+y}{2}=a.......\left( i \right)\]
We also know that the geometric mean (GM) of two numbers say a and b is given by \[\sqrt{ab}.\] So, we get the GM of its first term x and \[{{\left( 2n-1 \right)}^{th}}\] term y is given by \[\sqrt{xy}.\] We are given that \[{{n}^{th}}\] term of GP is b. So, we get,
\[GM=\sqrt{xy}=b......\left( ii \right)\]
We also know that the harmonic mean (HM) of two numbers say a and b is given by \[\dfrac{2ab}{a+b}.\] So, we get HM of its first term x and \[{{\left( 2n-1 \right)}^{th}}\] term y is given by \[\dfrac{2xy}{x+y}.\]
We are given that \[{{n}^{th}}\] term of HP is c. So, we get,
\[HM=\dfrac{2xy}{x+y}=c......\left( iii \right)\]
According to the rule of progression, we know that,
\[AM\ge GM\ge HM\]
So, by taking equations (i), (ii) and (iii), we get,
\[a\ge b\ge c\]
We also know that,
\[{{\left( GM \right)}^{2}}=\left( AM \right)\left( HM \right)\]
By substituting the values of AM, GM and HM from equation (i), (ii) and (iii), we get,
\[{{b}^{2}}=ac\]
Hence, options (b) and (d) are the right answers.

Note: In this question, students can verify \[{{b}^{2}}=ac\] by substituting the values of a, b and c terms of x and y as follows.
\[{{b}^{2}}=ac\]
By substituting \[b=\sqrt{xy},a=\dfrac{x+y}{2},c=\dfrac{2xy}{x+y}.\] We get,
\[{{\left( \sqrt{xy} \right)}^{2}}=\left( \dfrac{x+y}{2} \right)\left( \dfrac{2xy}{\left( x+y \right)} \right)\]
\[\Rightarrow xy=\dfrac{\left( x+y \right)}{\left( x+y \right)}\left( \dfrac{2xy}{2} \right)\]
\[\Rightarrow xy=xy\]
LHS = RHS
Hence, proved.
Students must note that AM = GM = HM when x = y, otherwise AM > GM > HM and we can verify this also by taking different values of x and y.