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# An A.P., a G.P. and H.P. have a and b for their first two terms. Show that their (n + 2) th terms will be in G.P. if $\dfrac{{{b}^{2n+2}}-{{a}^{2n+2}}}{ab\left( {{b}^{2n}}-{{a}^{2n}}\right)}=\dfrac{n+1}{n}$ . Verified
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Hint: First, find the (n + 2) th term of the A.P., G.P. and H.P. . Then apply the condition for them being in G.P. which is ${{y}^{2}}=xz$ if x, y, and z are in G.P. Using and simplifying this condition arrives at the required condition to prove.

Complete step by step solution:

We are given that an A.P., a G.P. and H.P. have a and b for their first two terms.
We need to show that their (n + 2) th terms will be in G.P. if $\dfrac{{{b}^{2n+2}}- {{a}^{2n+2}}}{ab\left( {{b}^{2n}}-{{a}^{2n}} \right)}=\dfrac{n+1}{n}$ .$\dfrac{{{b}^{2n+2}}- {{a}^{2n+2}}}{ab\left( {{b}^{2n}}-{{a}^{2n}} \right)}=\dfrac{n+1}{n}$
Let us have the common difference of the A.P. be d.
So, $d=b-a$
Let us have the common ratio of the G.P. be r.
So, $r=\dfrac{b}{a}$
Now, the (n+2) th term of the A.P. will be:
${{T}_{n+2}}=a+\left( n+1 \right)d=a+\left( n+1 \right)\left( b-a \right)$
${{T}_{n+2}}=-na+\left( n+1 \right)b$ for A.P. …(1)
Similarly, we will find the (n+2) th term of the G.P.

${{T}_{n+2}}=a{{r}^{n+1}}=a\cdot \dfrac{{{b}^{n+1}}}{{{a}^{n+1}}}$
${{T}_{n+2}}=\dfrac{{{b}^{n+1}}}{{{a}^{n}}}$ for G.P. …(2)
Now, we will find the (n+2) th term of the H.P.
For getting the (n+2) th term of the H.P., we will replace $a=\dfrac{1}{a}$ and
$b=\dfrac{1}{b}$ in equation (1). Then we will take the reciprocal of this resulting term.
First let us replace $a=\dfrac{1}{a}$ and $b=\dfrac{1}{b}$ in equation (1).
$-\dfrac{n}{a}+\dfrac{\left( n+1 \right)}{b}=\dfrac{\left( n+1 \right)a-bn}{ab}$
Now, we will take the reciprocal of this term to get the ${{T}_{n+2}}$ of the H.P.
${{T}_{n+2}}=\dfrac{ab}{\left( n+1 \right)a-bn}$ for H.P. …(3)
We are given that the above three terms in (1), (2) and (3) are themselves in G.P.

${{\left( \dfrac{{{b}^{n+1}}}{{{a}^{n}}} \right)}^{2}}=\left[ -na+\left( n+1 \right)b \right]\cdot \dfrac{ab}{\left( n+1 \right)a-bn}$
$\dfrac{{{b}^{2n+2}}}{{{a}^{2n}}}=\left[ -na+\left( n+1 \right)b \right]\cdot \dfrac{ab}{\left( n+1 \right)a-bn}$
$\dfrac{{{b}^{2n+1}}}{{{a}^{2n+1}}}=\dfrac{-na+\left( n+1 \right)b}{\left( n+1 \right)a-bn}$
$\left( n+1 \right)a{{b}^{2n+1}}-n\cdot {{b}^{2n+2}}=\left( n+1 \right)b{{a}^{2n+1}}-n\cdot {{a}^{2n+2}}$
$\left( n+1 \right)\left[ a{{b}^{2n+1}}-b{{a}^{2n+1}} \right]=n\left( {{b}^{2n+2}}-{{a}^{2n+2}} \right)$
Or, $\dfrac{{{b}^{2n+2}}-{{a}^{2n+2}}}{ab\left( {{b}^{2n}}-{{a}^{2n}} \right)}=\dfrac{n+1}{n}$
Hence proved.

Note: In this question, it is important to know the condition for three numbers being in a
G.P. This condition is if three numbers: x, y, and z are in G.P., then they will satisfy the
condition ${{y}^{2}}=xz$ .
Also, the calculations in this question are very lengthy and difficult. So approach them
carefully and try to avoid making any mistakes.
Last updated date: 25th Sep 2023
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