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Question: If the \[{{\left( p+q \right)}^{th}}\] term of a G.P. is A and \[{{\left( p-q \right)}^{th}}\] term ...

If the (p+q)th{{\left( p+q \right)}^{th}} term of a G.P. is A and (p−q)th{{\left( p-q \right)}^{th}} term is B, determine its pth{{p}^{th}} term.

Explanation

Solution

To find the value of value of pth{{p}^{th}} term, we will first write the (p+q)th{{\left( p+q \right)}^{th}} and (p−q)th{{\left( p-q \right)}^{th}} term of the GP. (p+q)th{{\left( p+q \right)}^{th}} term can be written as a×rp+q−1=Aa\times {{r}^{p+q-1}}=A and (p−q)th{{\left( p-q \right)}^{th}} term can be written as
a×rp−q−1=Ba\times {{r}^{p-q-1}}=B . Now, let us multiply these equations. After solving them we will get the pth{{p}^{th}} term.

Complete step by step answer:
We need to find the value of pth{{p}^{th}} term given that the (p+q)th{{\left( p+q \right)}^{th}} term of a G.P. is A and (p−q)th{{\left( p-q \right)}^{th}} term is B.
Let us recollect the representation of GP. A GP is represented as follows.
a,ar,ar2,...,arn−1a,ar,a{{r}^{2}},...,a{{r}^{n-1}} , where rr is the common ratio, aa is the first term, arar is the second,…, arn−1a{{r}^{n-1}} is the nth{{n}^{th}} term.
Hence, (p+q)th{{\left( p+q \right)}^{th}} term is represented as
a×rp+q−1=A...(i)a\times {{r}^{p+q-1}}=A...(i)
Similarly, we can find (p−q)th{{\left( p-q \right)}^{th}} term and is shown below.
a×rp−q−1=B...(ii)a\times {{r}^{p-q-1}}=B...(ii)
Now, let us multiply equations (i) and (ii), we will get
a×rp+q−1×a×rp−q−1=ABa\times {{r}^{p+q-1}}\times a\times {{r}^{p-q-1}}=AB
We know that an×am=am+n{{a}^{n}}\times {{a}^{m}}={{a}^{m+n}} . Hence, the above equation becomes
a2×rp+q−1+p−q−1=AB{{a}^{2}}\times {{r}^{p+q-1+p-q-1}}=AB
Solving the exponentials, we get
a2×r2p−2=AB{{a}^{2}}\times {{r}^{2p-2}}=AB
Taking 2 common from the exponent of rr , we will get
a2×r2(p−1)=AB{{a}^{2}}\times {{r}^{2(p-1)}}=AB
We know that (ab)m=ambm{{\left( ab \right)}^{m}}={{a}^{m}}{{b}^{m}} . Hence, the above equation can be written as
(ar(p−1))2=AB{{\left( a{{r}^{(p-1)}} \right)}^{2}}=AB
Now, let us take the square root. We will get
ar(p−1)=ABa{{r}^{(p-1)}}=\sqrt{AB}

Hence, the pth{{p}^{th}} term of the GP is AB\sqrt{AB} .

Note: GP is represented as a,ar,ar2,...,arn−1a,ar,a{{r}^{2}},...,a{{r}^{n-1}} . Without knowing this, further steps cannot be attained. Do not make mistakes in writing this equation. When multiplying the equations, be careful with the exponents. You must be thorough with the rules of exponents.