Question

If $ {a_1},{a_2},{a_3},...{a_n} $ are in H.P then $ {a_1}{a_2} + {a_2}{a_3} + ... + {a_{n - 1}}{a_n} = $

Answer 1

Hint: Recall that if $ {a_1},{a_2},{a_3},...{a_n} $ are in H.P that is Harmonic Progression, then the sequence \[\dfrac{1}{{{a_1}}},\dfrac{1}{{{a_2}}},...,\dfrac{1}{{{a_n}}}\] is in A.P that is Arithmetic Progression. Use the sum of such an A.P to find the relationship asked in the question.

Complete step-by-step answer:
We know that Harmonic Progression is formed by taking reciprocals of the terms of an Arithmetic Progression. From this we can see that if $ {a_1},{a_2},{a_3},...{a_n} $ are in H.P that is Harmonic Progression, then the sequence \[\dfrac{1}{{{a_1}}},\dfrac{1}{{{a_2}}},...,\dfrac{1}{{{a_n}}}\] is in A.P that is Arithmetic Progression. We will use this observation to find the value of the given expression.
In the question it is given that $ {a_1},{a_2},{a_3},...{a_n} $ are in H.P.
Therefore, \[\dfrac{1}{{{a_1}}},\dfrac{1}{{{a_2}}},\dfrac{1}{{{a_3}}}...,\dfrac{1}{{{a_n}}}\] are in A.P.
Let the common difference of the above mentioned A.P be $ d $ .
Then we know that \[d = \dfrac{1}{{{a_2}}} - \dfrac{1}{{{a_1}}}\].
 $ \Rightarrow d = \dfrac{{{a_1} - {a_2}}}{{{a_2}{a_1}}} $
 $ \Rightarrow {a_1}{a_2}d = {a_1} - {a_2} $ - - - - - - - - - - - - - (1)
Since, the difference between any two consecutive terms of an A.P is $ d $ , similarly we have,
 $ {a_2} - {a_3} = {a_2}{a_3}d $ - - - - - - - - - - - - - - (2)
Now we can continue this process and obtain $ {a_{n - 1}} - {a_n} = {a_{n - 1}}{a_n}d $ . - - - - - - - - - - (n-1)
Now we will add all these equations from (1) to (n-1) we get,
 $ {a_1} - {a_2} + {a_2} - {a_3} + ... + {a_{n - 2}} - {a_{n - 1}} + {a_{n - 1}} - {a_n} = {a_1}{a_2}d + {a_2}{a_3}d + ... + {a_n}{a_{n - 1}}d $
\[ \Rightarrow {a_1} - {a_n} = d\left( {{a_1}{a_2} + {a_2}{a_3} + ... + {a_n}{a_{n - 1}}} \right)\]
On diving the entire equation by d we get,
\[ \Rightarrow \dfrac{{{a_1} - {a_n}}}{d} = \left( {{a_1}{a_2} + {a_2}{a_3} + ... + {a_n}{a_{n - 1}}} \right)\] - - - - - - - - - - - - - - - - - - - (a)
Also, we know that $ \dfrac{1}{{{a_n}}} = \dfrac{1}{{{a_1}}} + (n - 1)d $ .
 $ \Rightarrow d = \dfrac{{{a_1} - {a_n}}}{{{a_1}{a_n}(n - 1)}} $
 $ \Rightarrow {a_1}{a_n}(n - 1) = \dfrac{{{a_1} - {a_n}}}{d} $
On substituting this value in (a) we get,
\[ \Rightarrow {a_1}{a_n}(n - 1) = \left( {{a_1}{a_2} + {a_2}{a_3} + ... + {a_n}{a_{n - 1}}} \right)\]
Hence $ {a_1}{a_2} + {a_2}{a_3} + ... + {a_{n - 1}}{a_n} = {a_1}{a_n}(n - 1) $ .


Note: Most of us stop once we get the equation (a) as we got a simplified version of the given expression. But you will lose marks for this as this answer contains a variable which is introduced by us that is d. So, we have to express it in terms of the variables given in the question.