Solveeit Logo
Login

Question

Question: If \({a_1}\),\({a_2}\) , …, \({a_{n - 1}}\) , \({a_n}\) are in A.P., then show that \(\dfrac{1}{{...

If a1{a_1},a2{a_2} , …, an−1{a_{n - 1}} , an{a_n} are in A.P., then show that
1a1a2+1a2an−1+1a3an−2+...+1ana1=2(a1+an)[1a1+1a2+...+1an]\dfrac{1}{{{a_1}{a_2}}} + \dfrac{1}{{{a_2}{a_{n - 1}}}} + \dfrac{1}{{{a_3}{a_{n - 2}}}} + ... + \dfrac{1}{{{a_n}{a_1}}} = \dfrac{2}{{\left( {{a_1} + {a_n}} \right)}}\left[ {\dfrac{1}{{{a_1}}} + \dfrac{1}{{{a_2}}} + ... + \dfrac{1}{{{a_n}}}} \right]

Explanation

Solution

In A.P we know that difference of two consecutive terms is constant. Then we know that the addition of first term and last term will be equal to the addition of second term and second last term and so on. Let as assume their addition to be equal to a constant K. Now take the first term on the left side and multiply by (a1+an)\left( {{a_1} + {a_n}} \right) on denominator and numerator. Solve and substitute the value a1+an=K{a_1} + {a_n} = Kin the formed function. Then follow this pattern for all the terms and add them and solve. In the last step again put a1+an=K{a_1} + {a_n} = K and the given sum will be equal to the RHS.

Complete step-by-step answer:
Given that a1{a_1},a2{a_2} , …, an−1{a_{n - 1}} , an{a_n} are in A.P. which means that that they have the same difference d between two consecutive terms.
It also means we can write-
⇒a2=a1+d\Rightarrow {a_2} = {a_1} + d , a3=a2+d{a_3} = {a_2} + d , …, an−1=an−2+d{a_{n - 1}} = {a_{n - 2}} + d , an=an−1+d{a_n} = {a_{n - 1}} + d
Also,
⇒a1an=a2an−1=a3an−2=...=K\Rightarrow {a_1}{a_n} = {a_2}{a_{n - 1}} = {a_3}{a_{n - 2}} = ... = K --- (i)(let’s say)
We have to prove-1a1a2+1a2an−1+1a3an−2+...+1ana1=2(a1+an)[1a1+1a2+...+1an]\dfrac{1}{{{a_1}{a_2}}} + \dfrac{1}{{{a_2}{a_{n - 1}}}} + \dfrac{1}{{{a_3}{a_{n - 2}}}} + ... + \dfrac{1}{{{a_n}{a_1}}} = \dfrac{2}{{\left( {{a_1} + {a_n}} \right)}}\left[ {\dfrac{1}{{{a_1}}} + \dfrac{1}{{{a_2}}} + ... + \dfrac{1}{{{a_n}}}} \right]
Now on taking LHS,
On multiplying the term (a1+an)\left( {{a_1} + {a_n}} \right) on denominator and numerator of the first term of LHS, we get
⇒1a1an=(a1+an)(a1+an)a1an\Rightarrow \dfrac{1}{{{a_1}{a_n}}} = \dfrac{{\left( {{a_1} + {a_n}} \right)}}{{\left( {{a_1} + {a_n}} \right){a_1}{a_n}}}
On solving we get,
\Rightarrow \dfrac{1}{{{a_1}{a_n}}} = \dfrac{1}{{{a_1} + {a_n}}}\left\\{ {\dfrac{{{a_1} + {a_n}}}{{{a_1}{a_n}}}} \right\\}
From eq. (i) we can write-
\Rightarrow \dfrac{1}{{{a_1}{a_n}}} = \dfrac{1}{K}\left\\{ {\dfrac{{{a_1} + {a_n}}}{{{a_1}{a_n}}}} \right\\}
Now we can write,
\Rightarrow \dfrac{1}{{{a_1}{a_n}}} = \dfrac{1}{K}\left\\{ {\dfrac{{{a_1}}}{{{a_1}{a_n}}} + \dfrac{{{a_n}}}{{{a_1}{a_n}}}} \right\\} = \dfrac{1}{K}\left\\{ {\dfrac{1}{{{a_n}}} + \dfrac{1}{{{a_1}}}} \right\\}
Following this pattern we can write second term as-
\Rightarrow \dfrac{1}{{{a_2}{a_{n - 1}}}} = \dfrac{1}{{{a_2} + {a_{n - 1}}}}\left\\{ {\dfrac{1}{{{a_{n - 1}}}} + \dfrac{1}{{{a_2}}}} \right\\}
Now using eq. (i) we can write all the terms as-
\Rightarrow \dfrac{1}{{{a_2}{a_{n - 1}}}} = \dfrac{1}{K}\left\\{ {\dfrac{1}{{{a_{n - 1}}}} + \dfrac{1}{{{a_2}}}} \right\\}
⇒\Rightarrow - - -- ‘’ --- ---
⇒\Rightarrow \dfrac{1}{{{a_n}{a_1}}} = \dfrac{1}{K}\left\\{ {\dfrac{1}{{{a_n}}} + \dfrac{1}{{{a_1}}}} \right\\}
Now on putting all these values in LHS we get,
⇒\Rightarrow \dfrac{1}{{{a_1}{a_2}}} + \dfrac{1}{{{a_2}{a_{n - 1}}}} + \dfrac{1}{{{a_3}{a_{n - 2}}}} + ... + \dfrac{1}{{{a_n}{a_1}}} = \dfrac{1}{K}\left\\{ {\dfrac{1}{{{a_n}}} + \dfrac{1}{{{a_1}}}} \right\\} + \dfrac{1}{K}\left\\{ {\dfrac{1}{{{a_{n - 1}}}} + \dfrac{1}{{{a_2}}}} \right\\} + ... + \dfrac{1}{K}\left\\{ {\dfrac{1}{{{a_n}}} + \dfrac{1}{{{a_1}}}} \right\\}
On taking 1K\dfrac{1}{K} common on the right side we get,
⇒\Rightarrow 1a1a2+1a2an−1+1a3an−2+...+1ana1=1K[1a1+1a2+...+1an+1a1+1a2+...++1an]\dfrac{1}{{{a_1}{a_2}}} + \dfrac{1}{{{a_2}{a_{n - 1}}}} + \dfrac{1}{{{a_3}{a_{n - 2}}}} + ... + \dfrac{1}{{{a_n}{a_1}}} = \dfrac{1}{K}\left[ {\dfrac{1}{{{a_1}}} + \dfrac{1}{{{a_2}}} + ... + \dfrac{1}{{{a_n}}} + \dfrac{1}{{{a_1}}} + \dfrac{1}{{{a_2}}} + ... + + \dfrac{1}{{{a_n}}}} \right]
Here we see that in right side there are two same terms of 1a1\dfrac{1}{{{a_1}}} to 1an\dfrac{1}{{{a_n}}} so on adding them we get,
⇒\Rightarrow 1a1a2+1a2an−1+1a3an−2+...+1ana1=1K[2a1+2a2+...+2an]\dfrac{1}{{{a_1}{a_2}}} + \dfrac{1}{{{a_2}{a_{n - 1}}}} + \dfrac{1}{{{a_3}{a_{n - 2}}}} + ... + \dfrac{1}{{{a_n}{a_1}}} = \dfrac{1}{K}\left[ {\dfrac{2}{{{a_1}}} + \dfrac{2}{{{a_2}}} + ... + \dfrac{2}{{{a_n}}}} \right] 1a1a2+1a2an−1+1a3an−2+...+1ana1=1K[2a1+2a2+...+2an]\dfrac{1}{{{a_1}{a_2}}} + \dfrac{1}{{{a_2}{a_{n - 1}}}} + \dfrac{1}{{{a_3}{a_{n - 2}}}} + ... + \dfrac{1}{{{a_n}{a_1}}} = \dfrac{1}{K}\left[ {\dfrac{2}{{{a_1}}} + \dfrac{2}{{{a_2}}} + ... + \dfrac{2}{{{a_n}}}} \right]
On taking the number 22 common on the right side we get,
⇒\Rightarrow 1a1a2+1a2an−1+1a3an−2+...+1ana1=2K[1a1+1a2+...+1an]\dfrac{1}{{{a_1}{a_2}}} + \dfrac{1}{{{a_2}{a_{n - 1}}}} + \dfrac{1}{{{a_3}{a_{n - 2}}}} + ... + \dfrac{1}{{{a_n}{a_1}}} = \dfrac{2}{K}\left[ {\dfrac{1}{{{a_1}}} + \dfrac{1}{{{a_2}}} + ... + \dfrac{1}{{{a_n}}}} \right]
Now from eq. (i) we know that a1+an=K{a_1} + {a_n} = K
So putting this value in above equation we get,
⇒\Rightarrow 1a1a2+1a2an−1+1a3an−2+...+1ana1=2(a1+an)[1a1+1a2+...+1an]\dfrac{1}{{{a_1}{a_2}}} + \dfrac{1}{{{a_2}{a_{n - 1}}}} + \dfrac{1}{{{a_3}{a_{n - 2}}}} + ... + \dfrac{1}{{{a_n}{a_1}}} = \dfrac{2}{{\left( {{a_1} + {a_n}} \right)}}\left[ {\dfrac{1}{{{a_1}}} + \dfrac{1}{{{a_2}}} + ... + \dfrac{1}{{{a_n}}}} \right]
So LHS=RHS
Hence Proved

Note: In A.P. series the behavior of the series depends on the value of the common difference as it can be negative or positive.
When the value of common difference is positive, the members of the series will go towards positive infinity.
When the value of common difference is negative, the members of the series will go towards negative infinity.