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# For a G.P, if ${{\left( m+n \right)}^{th}}$ term is p and ${{\left( m-n \right)}^{th}}$ term is q, then ${{m}^{th}}$ term is ________.(a) $pq$(b) $\sqrt{pq}$(c) $\dfrac{p}{q}$(d) $\dfrac{q}{p}$

Last updated date: 24th Jul 2024
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Hint: Write the formula for ${{\left( m+n \right)}^{th}}$ term and ${{\left( m-n \right)}^{th}}$ term and multiply them to get the expression for ${{m}^{th}}$ term directly.

We are given that ${{\left( m+n \right)}^{th}}$ term of G.P is p and ${{\left( m-n \right)}^{th}}$ term of G.P is q. We have to find the ${{m}^{th}}$ term of this G.P.
We know that any general geometric progression (G.P) series of the form
$a,ar,a{{r}^{2}},a{{r}^{3}}.....$
where a = first term, r = common ratio
We know that ${{n}^{th}}$ term of G.P is $a{{r}^{n-1}}$, i.e.
${{a}_{n}}=a{{r}^{n-1}}.....\left( i \right)$
Now, to get ${{\left( m+n \right)}^{th}}$ term, we will replace ‘n’ by $\left( m+n \right)$ in equation (i).
Therefore, we get ${{\left( m+n \right)}^{th}}$ term as
${{a}_{m+n}}=a{{r}^{m+n-1}}=p....\left( ii \right)$
Now, to get ${{\left( m-n \right)}^{th}}$ term, we will replace ‘n’ by $\left( m-n \right)$ in equation (i)
Therefore, we get ${{\left( m-n \right)}^{th}}$ term as
${{a}_{m-n}}=a{{r}^{m-n-1}}=q.....\left( iii \right)$
Now, by multiplying equation (ii) and (iii), we get,
$\left( {{a}_{m+n}} \right)\left( {{a}_{m-n}} \right)=\left( a{{r}^{m+n-1}} \right)\left( a{{r}^{m-n-1}} \right)=pq$
Since, we know that
${{a}^{m}}.{{a}^{n}}={{a}^{m+n}}$
Therefore, we get,
$\left( {{a}^{1+1}} \right).\left( {{r}^{\left( m+n-1 \right)+\left( m-n-1 \right)}} \right)=pq$
$=\left( {{a}^{2}} \right).{{r}^{\left( 2m-2 \right)}}=pq$
Or, ${{\left( a \right)}^{2}}{{\left( {{r}^{m-1}} \right)}^{2}}=pq$
By taking 2 common from powers, we get,
${{\left( a.{{r}^{m-1}} \right)}^{2}}=pq$
By taking square roots on both sides. We get,
$\sqrt{{{\left( a{{r}^{m-1}} \right)}^{2}}}=\sqrt{pq}$
Since, we know that
$\sqrt{{{a}^{2}}}=a$
Therefore, we get
$a.{{r}^{m-1}}=\sqrt{pq}....\left( iv \right)$
Now, to get the ${{m}^{th}}$ term, we will replace ‘n’ by ‘m’ in equation (i). We get
${{a}_{m}}=a{{r}^{m-1}}$
From equation (iv), we have found that $a{{r}^{m-1}}$ which is equal to ${{a}_{m}}\text{ or }{{m}^{th}}$ term. Therefore,
${{a}_{m}}=a{{r}^{m-1}}=\sqrt{pq}$
Hence, option (b) is correct.

Note: Some students make the mistake of finding ‘r’ and ‘a’ which are common ratio and first term of G.P separately and then putting it in the formula for ${{m}^{th}}$ term. But that is a time consuming and lengthy method as here we have found the ${{m}^{th}}$ term without finding ‘r’ and ‘a’ separately.