Question

If \[a,{{a}_{1}},{{a}_{2}},{{a}_{3}},...............,{{a}_{2n}},b\] are in AP and \[a,{{g}_{1}},{{g}_{2}},{{g}_{3}},..............,{{g}_{2n}},b\] are in GP and h is the harmonic mean of a and b then \[\dfrac{{{a}_{1}}+{{a}_{2n}}}{{{g}_{1}}{{g}_{2n}}}+\dfrac{{{a}_{2}}+{{a}_{2n-1}}}{{{g}_{2}}{{g}_{2n-1}}}+............+\dfrac{{{a}_{n}}+{{a}_{n+1}}}{{{g}_{n}}{{g}_{n+1}}}\] is equal to
(1) \[2nh\]
(2) \[\dfrac{n}{h}\]
(3) \[nh\]
(4) \[\dfrac{2n}{h}\]

Answer 1

Hint: The sum of the terms from the beginning and end of an AP will be equal and also it is equal to the sum of the first and last term. In GP the product of the terms from the beginning and end of the GP will be equal and also it is equal to the product of the first and last term. Keeping this in mind we will solve our problem.

Complete step-by-step solution:
AP stands for Arithmetic progression. An AP is a sequence in which each term except the first one differs from its previous term by a constant.
For example:
If we have three terms a,b, and c that are in AP, then
\[b-a=c-b\]
GP stands for Geometric progression. A GP is the sequence of terms in which all the succeeding terms will have a common ratio between them when they are divided by each other.
For example:
If a, b,c are the three terms that are in GP, then
\[\dfrac{b}{a}=\dfrac{c}{b}\]
HP stands for Harmonic Progression and it is obtained when we do the reciprocal of the terms which are in AP.
For example:
If we have three terms a,b, and c which are in AP then \[\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}\] will form a HP.
It is given in the question that
\[a,{{a}_{1}},{{a}_{2,}},{{a}_{3}},..............,{{a}_{2n}},b\] are in AP
\[a,{{g}_{1}},{{g}_{2}},{{g}_{3}},................,{{g}_{2n}},b\] are in GP
And also it is given that ‘h’ is the harmonic mean of a and b.
Now let us come back to our question,
If \[a,{{a}_{1}},{{a}_{2}},{{a}_{3}},................,{{a}_{2n}},b\] are in AP
Then,
 \[a+b={{a}_{1}}+{{a}_{2n}}\]
\[{{a}_{1}}+{{a}_{2n}}={{a}_{2}}+{{a}_{2n-1}}\]
\[\begin{align}
  & \Rightarrow a+b={{a}_{2}}+{{a}_{2n-1}} \\
 & \Rightarrow a+b={{a}_{n}}+{{a}_{n+1}} \\
\end{align}\] ……………..eq (1)
(As we know that the sum of the first and last term of an AP is constant)
If \[a,{{g}_{1}},{{g}_{2}},{{g}_{3}},...............,{{g}_{2n}},b\] are in GP
Then,
\[ab={{g}_{1}}{{g}_{2n}}\]
\[{{g}_{1}}{{g}_{2n}}={{g}_{2}}{{g}_{2n-1}}\]
\[\begin{align}
  & \Rightarrow ab={{g}_{2}}{{g}_{2n-1}} \\
 & \Rightarrow ab={{g}_{n}}{{g}_{n+1}} \\
\end{align}\] ……………….eq(2)
(As we know that the product of the first and last term of the GP is constant)
We know that, If ‘h’ is the harmonic mean of a and b then,
\[h=\dfrac{2ab}{a+b}\] ……………eq(3)
We have, \[\dfrac{{{a}_{1}}+{{a}_{2n}}}{{{g}_{1}}{{g}_{2n}}}+\dfrac{{{a}_{2}}+{{a}_{2n-1}}}{{{g}_{2}}{{g}_{2n-1}}}+...............+\dfrac{{{a}_{n}}+{{a}_{n+1}}}{{{g}_{n}}{{g}_{n+1}}}\]………..eq(4)
On putting the values of eq(1) and eq(2) in eq(4), we get
\[\dfrac{{{a}_{1}}+{{a}_{2n}}}{{{g}_{1}}{{g}_{2n}}}+\dfrac{{{a}_{2}}+{{a}_{2n-1}}}{{{g}_{2}}{{g}_{2n-1}}}+...........+\dfrac{{{a}_{n}}+{{a}_{n+1}}}{{{g}_{n}}{{g}_{n+1}}}\]
\[\Rightarrow \dfrac{a+b}{ab}+\dfrac{a+b}{ab}+...........+\dfrac{a+b}{ab}\] (up to n terms)
\[\Rightarrow \dfrac{{{a}_{1}}+{{a}_{2n}}}{{{g}_{1}}{{g}_{2n}}}+\dfrac{{{a}_{2}}+{{a}_{2n-1}}}{{{g}_{2}}{{g}_{2n-1}}}+...........+\dfrac{{{a}_{n}}+{{a}_{n+1}}}{{{g}_{n}}{{g}_{n+1}}}=n\left( \dfrac{a+b}{ab} \right)\]………..eq(5)
On multiplying the right-hand side of eq(5) by \[2\] in numerator and denominator, we get
\[\Rightarrow \dfrac{{{a}_{1}}+{{a}_{2n}}}{{{g}_{1}}{{g}_{2n}}}+\dfrac{{{a}_{2}}+{{a}_{2n-1}}}{{{g}_{2}}{{g}_{2n-1}}}+...........+\dfrac{{{a}_{n}}+{{a}_{n+1}}}{{{g}_{n}}{{g}_{n+1}}}=n\left( \dfrac{2(a+b)}{2ab} \right)\]………eq(6)
On dividing the right-hand side of eq(6) by \[a+b\] in numerator and denominator, we get
\[\begin{align}
  & \Rightarrow \dfrac{{{a}_{1}}+{{a}_{2n}}}{{{g}_{1}}{{g}_{2n}}}+\dfrac{{{a}_{2}}+{{a}_{2n-1}}}{{{g}_{2}}{{g}_{2n-1}}}+...........+\dfrac{{{a}_{n}}+{{a}_{n+1}}}{{{g}_{n}}{{g}_{n+1}}}=n\left( \dfrac{\dfrac{2(a+b)}{(a+b)}}{\dfrac{2ab}{(a+b)}} \right) \\
 & \\
\end{align}\]
\[\begin{align}
  & \Rightarrow \dfrac{{{a}_{1}}+{{a}_{2n}}}{{{g}_{1}}{{g}_{2n}}}+\dfrac{{{a}_{2}}+{{a}_{2n-1}}}{{{g}_{2}}{{g}_{2n-1}}}+...........+\dfrac{{{a}_{n}}+{{a}_{n+1}}}{{{g}_{n}}{{g}_{n+1}}}=\left( \dfrac{2n}{\dfrac{2ab}{(a+b)}} \right) \\
 & \\
\end{align}\]………….eq(7)
We know that,
\[h=\dfrac{2ab}{a+b}\] from eq (3)
So putting the value of ‘h’ in eq(7), we get
\[\Rightarrow \dfrac{{{a}_{1}}+{{a}_{2n}}}{{{g}_{1}}{{g}_{2n}}}+\dfrac{{{a}_{2}}+{{a}_{2n-1}}}{{{g}_{2}}{{g}_{2n-1}}}+...........+\dfrac{{{a}_{n}}+{{a}_{n+1}}}{{{g}_{n}}{{g}_{n+1}}}=\dfrac{2n}{h}\]
So the answer will be (4) \[\dfrac{2n}{h}\]

Note: If a constant is added to each of the terms of an AP then the resulting progression will also form an AP. If we do the reciprocal of each of the terms of the GP then the reciprocal terms will also form GP. Any of the terms of the HP cannot be equal to zero. A sequence will be in HP if every term will be the harmonic mean of its neighbouring term.