# If ${{x}_{1}},{{x}_{2}},.....,{{x}_{n}}\text{ and }\dfrac{1}{{{h}_{1}}},\dfrac{1}{{{h}_{2}}},.....,\dfrac{1}{{{h}_{n}}}$ are two A.P.s such that ${{x}_{3}}={{h}_{2}}=8\ and {{x}_{8}}={{h}_{7}}=20,$ then ${{x}_{5}}.{{h}_{10}}$ equals

(A) 2560

(B) 2650

(C) 3200

(D) 1600

Answer

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363.6k+ views

Hint: Use general term of AP, ${{T}_{n}}=a+\left( n-1 \right)d$ , where a is first term, d is common difference and n is number of terms. Find the first term and common difference of the given series to calculate ${{x}_{5}}\And {{h}_{10}}$.

We have $\left( {{x}_{1}},{{x}_{2}},.....,{{x}_{n}} \right)\text{ and }\left( \dfrac{1}{{{h}_{1}}},\dfrac{1}{{{h}_{2}}},.....,\dfrac{1}{{{h}_{n}}} \right)$ are two AP’s where,

${{x}_{3}}={{h}_{2}}=8$………………… (1)

${{x}_{8}}={{h}_{7}}=20$……………… (2)

We need to determine ${{x}_{5}}.{{h}_{10}}=?$

As we know general term of an A.P, where first term is ‘a’ and common difference is ‘d’

${{T}_{n}}=a+\left( n-1 \right)d$………………. (3)

Now, from equation (1), we have

${{x}_{3}}={{h}_{2}}=8$

From equation (3), we can calculate ${{x}_{3}}$ as;

Let $\left( {{x}_{1}},{{x}_{2}},.....,{{x}_{n}} \right)$ have common difference of ${{d}_{1}}$ then,

Hence,

$\begin{align}

& {{x}_{3}}={{x}_{1}}+\left( 3-1 \right){{d}_{1}} \\

& {{x}_{3}}={{x}_{1}}+2{{d}_{1}}=8 \\

\end{align}$ …………(4)

For ${{h}_{2}}$, we have given that $\dfrac{1}{{{h}_{1}}},\dfrac{1}{{{h}_{2}}},.....,\dfrac{1}{{{h}_{n}}}$are in AP, let common difference of this AP be ${{d}_{2}}$, therefore we can write

$\begin{align}

& \dfrac{1}{{{h}_{2}}}=\dfrac{1}{{{h}_{1}}}+\left( 2-1 \right){{d}_{2}} \\

& \dfrac{1}{{{h}_{2}}}=\dfrac{1}{{{h}_{1}}}+\dfrac{{{d}_{2}}}{1}=\dfrac{1+{{h}_{1}}{{d}_{2}}}{{{h}_{1}}} \\

& {{h}_{2}}=\dfrac{{{h}_{1}}}{{{h}_{1}}{{d}_{2}}+1}=8 \\

& {{h}_{1}}=8{{h}_{1}}{{d}_{2}}+8 \\

& {{h}_{1}}\left( 1-8{{d}_{2}} \right)=8 \\

\end{align}$

${{h}_{1}}=\dfrac{8}{1-8{{d}_{2}}}$…………………. (5)

Now, from the equation (2) we have

${{x}_{8}}={{h}_{7}}=20$

${{x}_{8}}$can be written by equation (3). As, we have ${{x}_{1}},{{x}_{2}},.....,{{x}_{n}}$are in AP with c.d.;

${{x}_{8}}={{x}_{1}}+\left( 8-1 \right){{d}_{1}}=20$

${{x}_{1}}+7{{d}_{1}}=20$……………….. (6)

For ${{h}_{7}}$ we have series $\dfrac{1}{{{h}_{1}}},\dfrac{1}{{{h}_{2}}},.....,\dfrac{1}{{{h}_{n}}}$in AP with $c.d={{d}_{2}}$ from there, we can write;

\[\begin{align}

& \dfrac{1}{{{h}_{7}}}=\dfrac{1}{{{h}_{1}}}+6{{d}_{2}} \\

& \dfrac{1}{{{h}_{7}}}=\dfrac{1+6{{d}_{2}}{{h}_{1}}}{{{h}_{1}}} \\

& {{h}_{7}}=\dfrac{{{h}_{1}}}{1+6{{d}_{2}}{{h}_{1}}}=20 \\

& {{h}_{1}}=20+120{{d}_{2}}{{h}_{1}} \\

\end{align}\]

\[{{h}_{1}}=\dfrac{20}{1-120{{d}_{2}}}\]…………………. (7)

Now, we can get ${{x}_{1}}\And {{d}_{1}}$ from equation (4) and (6) by solving them as follows;

Therefore, we have;

${{x}_{1}}+2{{d}_{1}}=8\And {{x}_{1}}+7{{d}_{1}}=20$

Subtracting both the equations, we get

\[\dfrac{\begin{matrix}

{{x}_{1}}+2{{d}_{1}}=8 \\

\begin{align}

& {{x}_{1}}+7{{d}_{1}}=20 \\

& - \\

\end{align} \\

\end{matrix}}{0-5{{d}_{1}}=-12}\]

Hence,${{d}_{1}}=\dfrac{12}{5}$

Now, for ${{x}_{1}}$, we have ${{x}_{1}}+2{{d}_{1}}=8$

$\begin{align}

& \because {{x}_{1}}+2\times \dfrac{12}{5}=8 \\

& {{x}_{1}}=8-\dfrac{24}{5}=\dfrac{16}{5} \\

\end{align}$

Hence, we get

${{x}_{1}}=\dfrac{16}{5},{{d}_{1}}=\dfrac{12}{5}$ ……………….. (8)

For ${{h}_{1}}\And {{d}_{2}}$, we have equations (5) and (7), we get

\[{{h}_{1}}=\dfrac{8}{1-8{{d}_{2}}}=\dfrac{20}{1-120{{d}_{2}}}\]

On simplifying above relation, we get;

\[\dfrac{2}{1-8{{d}_{2}}}=\dfrac{5}{1-120{{d}_{2}}}\]

On cross – multiplying, we get

$\begin{align}

& 2-240{{d}_{2}}=5-40{{d}_{2}} \\

& -3=200{{d}_{2}} \\

& {{d}_{2}}=\dfrac{-3}{200} \\

\end{align}$

Now, for calculating ${{h}_{1}}$, we have;

\[\begin{align}

& {{h}_{1}}=\dfrac{8}{1-8{{d}_{2}}}=\dfrac{8}{1-8\times \left( \dfrac{-3}{200} \right)} \\

& {{h}_{1}}=\dfrac{8}{1+\left( \dfrac{3}{25} \right)}=\dfrac{8\times 25}{28} \\

& {{h}_{1}}=\dfrac{50}{7} \\

\end{align}\]

Hence, Now we have;

\[{{h}_{1}}=\dfrac{50}{7},{{d}_{2}}=\dfrac{-3}{200}\]……………….. (9)

Now, coming to question we have to calculate

${{x}_{5}}={{h}_{10}}$

For ${{x}_{5}}$, we have first term of series ${{x}_{1}}=\dfrac{16}{5}$ and common difference ${{d}_{1}}=\dfrac{12}{5}$ from equation (8).

Hence, from equation (3), we have general formula for A.P., we get

$\begin{align}

& {{x}_{5}}={{x}_{1}}+4{{d}_{1}} \\

& {{x}_{5}}=\dfrac{16}{5}+4\times \dfrac{12}{5}=\dfrac{64}{5}...........\left( 10 \right) \\

\end{align}$

For ${{h}_{10}}$, we have series $\dfrac{1}{{{h}_{1}}},\dfrac{1}{{{h}_{2}}},.....,\dfrac{1}{{{h}_{n}}}$ in A.P and \[{{h}_{1}}=\dfrac{50}{7}\] and common difference \[{{d}_{2}}=\dfrac{-3}{200}\] from equation (9).

Using equation (5), we can write;

$\begin{align}

& \dfrac{1}{{{h}_{10}}}=\dfrac{1}{{{h}_{1}}}+9{{d}_{2}}=\dfrac{7}{50}+9\times \dfrac{-3}{200} \\

& \dfrac{1}{{{h}_{10}}}=\dfrac{7}{50}-\dfrac{27}{200}=\dfrac{28-27}{200} \\

& {{h}_{10}}=200.....................................\left( 11 \right) \\

\end{align}$

Now, ${{x}_{5}}.{{h}_{10}}$ can be calculated from equations (10) and (11), we get;

$\begin{align}

& {{x}_{5}}.{{h}_{10}}=\dfrac{64}{5}\times 200=64\times 40 \\

& {{x}_{5}}.{{h}_{10}}=2560 \\

\end{align}$

Hence, the answer is option (A).

Note: One can be easily confused with second series and can write general term for this series as

$\begin{align}

& {{T}_{n}}={{h}_{1}}+\left( n-1 \right)d \\

& Or \\

& \dfrac{1}{{{h}_{1}}}+\left( n-1 \right)\dfrac{1}{{{d}_{1}}} \\

\end{align}$

Which will be wrong and give the wrong solution. Correct general term for the second series is

${{T}_{n}}=\dfrac{1}{{{h}_{1}}}+\left( n-1 \right)d$.

__Complete step-by-step answer:__We have $\left( {{x}_{1}},{{x}_{2}},.....,{{x}_{n}} \right)\text{ and }\left( \dfrac{1}{{{h}_{1}}},\dfrac{1}{{{h}_{2}}},.....,\dfrac{1}{{{h}_{n}}} \right)$ are two AP’s where,

${{x}_{3}}={{h}_{2}}=8$………………… (1)

${{x}_{8}}={{h}_{7}}=20$……………… (2)

We need to determine ${{x}_{5}}.{{h}_{10}}=?$

As we know general term of an A.P, where first term is ‘a’ and common difference is ‘d’

${{T}_{n}}=a+\left( n-1 \right)d$………………. (3)

Now, from equation (1), we have

${{x}_{3}}={{h}_{2}}=8$

From equation (3), we can calculate ${{x}_{3}}$ as;

Let $\left( {{x}_{1}},{{x}_{2}},.....,{{x}_{n}} \right)$ have common difference of ${{d}_{1}}$ then,

Hence,

$\begin{align}

& {{x}_{3}}={{x}_{1}}+\left( 3-1 \right){{d}_{1}} \\

& {{x}_{3}}={{x}_{1}}+2{{d}_{1}}=8 \\

\end{align}$ …………(4)

For ${{h}_{2}}$, we have given that $\dfrac{1}{{{h}_{1}}},\dfrac{1}{{{h}_{2}}},.....,\dfrac{1}{{{h}_{n}}}$are in AP, let common difference of this AP be ${{d}_{2}}$, therefore we can write

$\begin{align}

& \dfrac{1}{{{h}_{2}}}=\dfrac{1}{{{h}_{1}}}+\left( 2-1 \right){{d}_{2}} \\

& \dfrac{1}{{{h}_{2}}}=\dfrac{1}{{{h}_{1}}}+\dfrac{{{d}_{2}}}{1}=\dfrac{1+{{h}_{1}}{{d}_{2}}}{{{h}_{1}}} \\

& {{h}_{2}}=\dfrac{{{h}_{1}}}{{{h}_{1}}{{d}_{2}}+1}=8 \\

& {{h}_{1}}=8{{h}_{1}}{{d}_{2}}+8 \\

& {{h}_{1}}\left( 1-8{{d}_{2}} \right)=8 \\

\end{align}$

${{h}_{1}}=\dfrac{8}{1-8{{d}_{2}}}$…………………. (5)

Now, from the equation (2) we have

${{x}_{8}}={{h}_{7}}=20$

${{x}_{8}}$can be written by equation (3). As, we have ${{x}_{1}},{{x}_{2}},.....,{{x}_{n}}$are in AP with c.d.;

${{x}_{8}}={{x}_{1}}+\left( 8-1 \right){{d}_{1}}=20$

${{x}_{1}}+7{{d}_{1}}=20$……………….. (6)

For ${{h}_{7}}$ we have series $\dfrac{1}{{{h}_{1}}},\dfrac{1}{{{h}_{2}}},.....,\dfrac{1}{{{h}_{n}}}$in AP with $c.d={{d}_{2}}$ from there, we can write;

\[\begin{align}

& \dfrac{1}{{{h}_{7}}}=\dfrac{1}{{{h}_{1}}}+6{{d}_{2}} \\

& \dfrac{1}{{{h}_{7}}}=\dfrac{1+6{{d}_{2}}{{h}_{1}}}{{{h}_{1}}} \\

& {{h}_{7}}=\dfrac{{{h}_{1}}}{1+6{{d}_{2}}{{h}_{1}}}=20 \\

& {{h}_{1}}=20+120{{d}_{2}}{{h}_{1}} \\

\end{align}\]

\[{{h}_{1}}=\dfrac{20}{1-120{{d}_{2}}}\]…………………. (7)

Now, we can get ${{x}_{1}}\And {{d}_{1}}$ from equation (4) and (6) by solving them as follows;

Therefore, we have;

${{x}_{1}}+2{{d}_{1}}=8\And {{x}_{1}}+7{{d}_{1}}=20$

Subtracting both the equations, we get

\[\dfrac{\begin{matrix}

{{x}_{1}}+2{{d}_{1}}=8 \\

\begin{align}

& {{x}_{1}}+7{{d}_{1}}=20 \\

& - \\

\end{align} \\

\end{matrix}}{0-5{{d}_{1}}=-12}\]

Hence,${{d}_{1}}=\dfrac{12}{5}$

Now, for ${{x}_{1}}$, we have ${{x}_{1}}+2{{d}_{1}}=8$

$\begin{align}

& \because {{x}_{1}}+2\times \dfrac{12}{5}=8 \\

& {{x}_{1}}=8-\dfrac{24}{5}=\dfrac{16}{5} \\

\end{align}$

Hence, we get

${{x}_{1}}=\dfrac{16}{5},{{d}_{1}}=\dfrac{12}{5}$ ……………….. (8)

For ${{h}_{1}}\And {{d}_{2}}$, we have equations (5) and (7), we get

\[{{h}_{1}}=\dfrac{8}{1-8{{d}_{2}}}=\dfrac{20}{1-120{{d}_{2}}}\]

On simplifying above relation, we get;

\[\dfrac{2}{1-8{{d}_{2}}}=\dfrac{5}{1-120{{d}_{2}}}\]

On cross – multiplying, we get

$\begin{align}

& 2-240{{d}_{2}}=5-40{{d}_{2}} \\

& -3=200{{d}_{2}} \\

& {{d}_{2}}=\dfrac{-3}{200} \\

\end{align}$

Now, for calculating ${{h}_{1}}$, we have;

\[\begin{align}

& {{h}_{1}}=\dfrac{8}{1-8{{d}_{2}}}=\dfrac{8}{1-8\times \left( \dfrac{-3}{200} \right)} \\

& {{h}_{1}}=\dfrac{8}{1+\left( \dfrac{3}{25} \right)}=\dfrac{8\times 25}{28} \\

& {{h}_{1}}=\dfrac{50}{7} \\

\end{align}\]

Hence, Now we have;

\[{{h}_{1}}=\dfrac{50}{7},{{d}_{2}}=\dfrac{-3}{200}\]……………….. (9)

Now, coming to question we have to calculate

${{x}_{5}}={{h}_{10}}$

For ${{x}_{5}}$, we have first term of series ${{x}_{1}}=\dfrac{16}{5}$ and common difference ${{d}_{1}}=\dfrac{12}{5}$ from equation (8).

Hence, from equation (3), we have general formula for A.P., we get

$\begin{align}

& {{x}_{5}}={{x}_{1}}+4{{d}_{1}} \\

& {{x}_{5}}=\dfrac{16}{5}+4\times \dfrac{12}{5}=\dfrac{64}{5}...........\left( 10 \right) \\

\end{align}$

For ${{h}_{10}}$, we have series $\dfrac{1}{{{h}_{1}}},\dfrac{1}{{{h}_{2}}},.....,\dfrac{1}{{{h}_{n}}}$ in A.P and \[{{h}_{1}}=\dfrac{50}{7}\] and common difference \[{{d}_{2}}=\dfrac{-3}{200}\] from equation (9).

Using equation (5), we can write;

$\begin{align}

& \dfrac{1}{{{h}_{10}}}=\dfrac{1}{{{h}_{1}}}+9{{d}_{2}}=\dfrac{7}{50}+9\times \dfrac{-3}{200} \\

& \dfrac{1}{{{h}_{10}}}=\dfrac{7}{50}-\dfrac{27}{200}=\dfrac{28-27}{200} \\

& {{h}_{10}}=200.....................................\left( 11 \right) \\

\end{align}$

Now, ${{x}_{5}}.{{h}_{10}}$ can be calculated from equations (10) and (11), we get;

$\begin{align}

& {{x}_{5}}.{{h}_{10}}=\dfrac{64}{5}\times 200=64\times 40 \\

& {{x}_{5}}.{{h}_{10}}=2560 \\

\end{align}$

Hence, the answer is option (A).

Note: One can be easily confused with second series and can write general term for this series as

$\begin{align}

& {{T}_{n}}={{h}_{1}}+\left( n-1 \right)d \\

& Or \\

& \dfrac{1}{{{h}_{1}}}+\left( n-1 \right)\dfrac{1}{{{d}_{1}}} \\

\end{align}$

Which will be wrong and give the wrong solution. Correct general term for the second series is

${{T}_{n}}=\dfrac{1}{{{h}_{1}}}+\left( n-1 \right)d$.

Last updated date: 01st Oct 2023

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