# 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: 26th Mar 2023

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Answer

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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.

__Complete step-by-step answer:__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.

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