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Question: If \({{a}_{1}},{{a}_{2}},{{a}_{3}}.......{{a}_{n}}\) are in A.P., where \({{a}_{i}}>0\) for all \(i\...

If a1,a2,a3.......an{{a}_{1}},{{a}_{2}},{{a}_{3}}.......{{a}_{n}} are in A.P., where ai>0{{a}_{i}}>0 for all ii, then the value of 1a1+a2+1a2+a3+....+1an−1+an=\dfrac{1}{\sqrt{{{a}_{1}}}+\sqrt{{{a}_{2}}}}+\dfrac{1}{\sqrt{{{a}_{2}}}+\sqrt{{{a}_{3}}}}+....+\dfrac{1}{\sqrt{{{a}_{n-1}}}+\sqrt{{{a}_{n}}}}=
A. n−1a1+an\dfrac{n-1}{\sqrt{{{a}_{1}}}+\sqrt{{{a}_{n}}}}
B. n+1a1+an\dfrac{n+1}{\sqrt{{{a}_{1}}}+\sqrt{{{a}_{n}}}}
C. n−1a1−an\dfrac{n-1}{\sqrt{{{a}_{1}}}-\sqrt{{{a}_{n}}}}
D. n+1a1−an\dfrac{n+1}{\sqrt{{{a}_{1}}}-\sqrt{{{a}_{n}}}}.

Explanation

Solution

Hint:We will be using the concept of series and progression to solve the problem. We will first rationalize the terms of the series we have to find then we will use the value of difference of two consecutive terms of an A.P that is a common difference to further simplify the problem.

Complete step-by-step answer:
Now, we have been given that a1,a2,a3.......an{{a}_{1}},{{a}_{2}},{{a}_{3}}.......{{a}_{n}} are in A.P.
Now, we have to find the value of 1a1+a2+1a2+a3+....+1an−1+an\dfrac{1}{\sqrt{{{a}_{1}}}+\sqrt{{{a}_{2}}}}+\dfrac{1}{\sqrt{{{a}_{2}}}+\sqrt{{{a}_{3}}}}+....+\dfrac{1}{\sqrt{{{a}_{n-1}}}+\sqrt{{{a}_{n}}}}.
Now, we will rationalize each term of the series with the conjugate of their surd in the denominator because doing so we will have the square root removed in the denominator and since we can see that each term of the series have consecutive terms of an A.P and by doing rationalization we will have the difference of the consecutive terms and we know that the difference of the consecutive terms of an AP is same therefore we will have denominator in each term of the series same . So, we have the series as,
a1−a2(a1+a2)(a1−a2)+(a2−a3)(a2+a3)(a2−a3)+....+an−1−anan−1+an\dfrac{\sqrt{{{a}_{1}}}-\sqrt{{{a}_{2}}}}{\left( \sqrt{{{a}_{1}}}+\sqrt{{{a}_{2}}} \right)\left( \sqrt{{{a}_{1}}}-\sqrt{{{a}_{2}}} \right)}+\dfrac{\left( \sqrt{{{a}_{2}}}-\sqrt{{{a}_{3}}} \right)}{\left( \sqrt{{{a}_{2}}}+\sqrt{{{a}_{3}}} \right)\left( \sqrt{{{a}_{2}}}-\sqrt{{{a}_{3}}} \right)}+....+\dfrac{\sqrt{{{a}_{n-1}}}-\sqrt{{{a}_{n}}}}{\sqrt{{{a}_{n-1}}}+\sqrt{{{a}_{n}}}}
Now, we know the trigonometric identity that,
(a−b)(a+b)=a−b\left( \sqrt{a}-\sqrt{b} \right)\left( \sqrt{a}+\sqrt{b} \right)=a-b
So, we have the series as,
a1−a2a1−a2+a2−a3a2−a3+....+an−1−anan−1−an\dfrac{\sqrt{{{a}_{1}}}-\sqrt{{{a}_{2}}}}{{{a}_{1}}-{{a}_{2}}}+\dfrac{\sqrt{{{a}_{2}}}-\sqrt{{{a}_{3}}}}{{{a}_{2}}-{{a}_{3}}}+....+\dfrac{\sqrt{{{a}_{n-1}}}-\sqrt{{{a}_{n}}}}{{{a}_{n-1}}-{{a}_{n}}}
Now, we know that the difference between two consecutive terms in A.P is the same. So, we have,
a1−a2=a2−a3=a3−a4.........=an−1−a4..........(1){{a}_{1}}-{{a}_{2}}={{a}_{2}}-{{a}_{3}}={{a}_{3}}-{{a}_{4}}.........={{a}_{n-1}}-{{a}_{4}}..........\left( 1 \right)
Also, we know the general nth term of an A.P with first term a and common difference d is,
Tn=a+(n−1)d{{T}_{n}}=a+\left( n-1 \right)d
Now, the series becomes,
\dfrac{1}{{{a}_{1}}-{{a}_{2}}}\left\\{ \sqrt{{{a}_{1}}}-\sqrt{{{a}_{2}}}+\sqrt{{{a}_{2}}}-\sqrt{{{a}_{3}}}+.....\sqrt{{{a}_{n-1}}}-\sqrt{{{a}_{n}}} \right\\}
We have taken the denominator in all the terms of sum as a1−a2{{a}_{1}}-{{a}_{2}} from (1).
Now, we can see that all the terms are cancelling each other except the first and the last term. So, we have,
\dfrac{1}{{{a}_{1}}-{{a}_{2}}}\left\\{ \sqrt{{{a}_{1}}}-\sqrt{{{a}_{n}}} \right\\}
Now, we will again rationalize it. So, we have,
=(a1−an)(a1+an)(a1−a2)(a1+an) =(a1−an)(a1−a2)(a1+an) \begin{aligned} & =\dfrac{\left( \sqrt{{{a}_{1}}}-\sqrt{{{a}_{n}}} \right)\left( \sqrt{{{a}_{1}}}+\sqrt{{{a}_{n}}} \right)}{\left( {{a}_{1}}-{{a}_{2}} \right)\left( \sqrt{{{a}_{1}}}+\sqrt{{{a}_{n}}} \right)} \\\ & =\dfrac{\left( {{a}_{1}}-{{a}_{n}} \right)}{\left( {{a}_{1}}-{{a}_{2}} \right)\left( \sqrt{{{a}_{1}}}+\sqrt{{{a}_{n}}} \right)} \\\ \end{aligned}
Now, we know that the general term of an A.P i.e. nth term is,
Tn=a+(n−1)d{{T}_{n}}=a+\left( n-1 \right)d
Now, for this A.P, we have,
T2=a1+(2−1)d=a2 Tn=a1+(n−1)d=an \begin{aligned} & {{T}_{2}}={{a}_{1}}+\left( 2-1 \right)d={{a}_{2}} \\\ & {{T}_{n}}={{a}_{1}}+\left( n-1 \right)d={{a}_{n}} \\\ \end{aligned}
So, we have the sum as,
=(a1−(a1+(n−1)d))(a1−(a1+d))(a1+an) =(a1−a1−(n−1)d)(a1−a1−d)(a1+an) =−(n−1)d−d(a1+an) \begin{aligned} & =\dfrac{\left( {{a}_{1}}-\left( {{a}_{1}}+\left( n-1 \right)d \right) \right)}{\left( {{a}_{1}}-\left( {{a}_{1}}+d \right) \right)\left( \sqrt{{{a}_{1}}}+\sqrt{{{a}_{n}}} \right)} \\\ & =\dfrac{\left( {{a}_{1}}-{{a}_{1}}-\left( n-1 \right)d \right)}{\left( {{a}_{1}}-{{a}_{1}}-d \right)\left( \sqrt{{{a}_{1}}}+\sqrt{{{a}_{n}}} \right)} \\\ & =\dfrac{-\left( n-1 \right)d}{-d\left( \sqrt{{{a}_{1}}}+\sqrt{{{a}_{n}}} \right)} \\\ \end{aligned}
Now, we will cancel –d in both numerator and denominator. So, we have,
n−1a1+an\dfrac{n-1}{\sqrt{{{a}_{1}}}+\sqrt{{{a}_{n}}}}
Hence, the correct option is (A).

Note: To solve these types of questions it is important to note that we have used the fact that the difference between two consecutive terms of an A.P. is the same. Also, we have to note that we have cancelled the terms in consecutive ways so that we are left with the first and last term. This is a unique trick to solve these types of questions.