
If the \[{{\left( p+q \right)}^{th}}\] term of a G.P. is A and \[{{\left( p-q \right)}^{th}}\] term is B, determine its ${{p}^{th}}$ term.
Answer
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Hint: To find the value of value of ${{p}^{th}}$ term, we will first write the \[{{\left( p+q \right)}^{th}}\] and \[{{\left( p-q \right)}^{th}}\] term of the GP. \[{{\left( p+q \right)}^{th}}\] term can be written as \[a\times {{r}^{p+q-1}}=A\] and \[{{\left( p-q \right)}^{th}}\] term can be written as
$a\times {{r}^{p-q-1}}=B$ . Now, let us multiply these equations. After solving them we will get the ${{p}^{th}}$ term.
Complete step by step answer:
We need to find the value of ${{p}^{th}}$ term given that the \[{{\left( p+q \right)}^{th}}\] term of a G.P. is A and \[{{\left( p-q \right)}^{th}}\] term is B.
Let us recollect the representation of GP. A GP is represented as follows.
$a,ar,a{{r}^{2}},...,a{{r}^{n-1}}$ , where $r$ is the common ratio, $a$ is the first term, $ar$ is the second,…, $a{{r}^{n-1}}$ is the ${{n}^{th}}$ term.
Hence, \[{{\left( p+q \right)}^{th}}\] term is represented as
$a\times {{r}^{p+q-1}}=A...(i)$
Similarly, we can find \[{{\left( p-q \right)}^{th}}\] term and is shown below.
$a\times {{r}^{p-q-1}}=B...(ii)$
Now, let us multiply equations (i) and (ii), we will get
$a\times {{r}^{p+q-1}}\times a\times {{r}^{p-q-1}}=AB$
We know that ${{a}^{n}}\times {{a}^{m}}={{a}^{m+n}}$ . Hence, the above equation becomes
${{a}^{2}}\times {{r}^{p+q-1+p-q-1}}=AB$
Solving the exponentials, we get
${{a}^{2}}\times {{r}^{2p-2}}=AB$
Taking 2 common from the exponent of $r$ , we will get
${{a}^{2}}\times {{r}^{2(p-1)}}=AB$
We know that ${{\left( ab \right)}^{m}}={{a}^{m}}{{b}^{m}}$ . Hence, the above equation can be written as
${{\left( a{{r}^{(p-1)}} \right)}^{2}}=AB$
Now, let us take the square root. We will get
$a{{r}^{(p-1)}}=\sqrt{AB}$
Hence, the ${{p}^{th}}$ term of the GP is $\sqrt{AB}$ .
Note: GP is represented as $a,ar,a{{r}^{2}},...,a{{r}^{n-1}}$ . 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.
$a\times {{r}^{p-q-1}}=B$ . Now, let us multiply these equations. After solving them we will get the ${{p}^{th}}$ term.
Complete step by step answer:
We need to find the value of ${{p}^{th}}$ term given that the \[{{\left( p+q \right)}^{th}}\] term of a G.P. is A and \[{{\left( p-q \right)}^{th}}\] term is B.
Let us recollect the representation of GP. A GP is represented as follows.
$a,ar,a{{r}^{2}},...,a{{r}^{n-1}}$ , where $r$ is the common ratio, $a$ is the first term, $ar$ is the second,…, $a{{r}^{n-1}}$ is the ${{n}^{th}}$ term.
Hence, \[{{\left( p+q \right)}^{th}}\] term is represented as
$a\times {{r}^{p+q-1}}=A...(i)$
Similarly, we can find \[{{\left( p-q \right)}^{th}}\] term and is shown below.
$a\times {{r}^{p-q-1}}=B...(ii)$
Now, let us multiply equations (i) and (ii), we will get
$a\times {{r}^{p+q-1}}\times a\times {{r}^{p-q-1}}=AB$
We know that ${{a}^{n}}\times {{a}^{m}}={{a}^{m+n}}$ . Hence, the above equation becomes
${{a}^{2}}\times {{r}^{p+q-1+p-q-1}}=AB$
Solving the exponentials, we get
${{a}^{2}}\times {{r}^{2p-2}}=AB$
Taking 2 common from the exponent of $r$ , we will get
${{a}^{2}}\times {{r}^{2(p-1)}}=AB$
We know that ${{\left( ab \right)}^{m}}={{a}^{m}}{{b}^{m}}$ . Hence, the above equation can be written as
${{\left( a{{r}^{(p-1)}} \right)}^{2}}=AB$
Now, let us take the square root. We will get
$a{{r}^{(p-1)}}=\sqrt{AB}$
Hence, the ${{p}^{th}}$ term of the GP is $\sqrt{AB}$ .
Note: GP is represented as $a,ar,a{{r}^{2}},...,a{{r}^{n-1}}$ . 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.
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