Question

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 1

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 ${{x}_{5}}\And {{h}_{10}}$.

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