Question

If ${{a}_{1}},{{a}_{2}},{{a}_{3}},....{{a}_{n}}$ are in A.P. where ${{a}_{i}}>0$ for all i then $\dfrac{1}{\sqrt{{{a}_{1}}}+\sqrt{{{a}_{2}}}}+\dfrac{1}{\sqrt{{{a}_{2}}}+\sqrt{{{a}_{3}}}}+......\dfrac{1}{\sqrt{{{a}_{n-1}}}+\sqrt{{{a}_{n}}}}$ equals to
(A) $\dfrac{1}{\left( n-1 \right)\left( \sqrt{{{a}_{1}}}+\sqrt{{{a}_{n}}} \right)}$
(B) $\dfrac{n-1}{\left( \sqrt{{{a}_{1}}}\times \sqrt{{{a}_{n}}} \right)}$
(C) $\dfrac{1}{\left( \sqrt{{{a}_{1}}}+\sqrt{{{a}_{n}}} \right)}$
(D) $\dfrac{n-1}{\left( \sqrt{{{a}_{1}}}+\sqrt{{{a}_{n}}} \right)}$

Answer 1

From the question we have been given that ${{a}_{1}},{{a}_{2}},{{a}_{3}},....{{a}_{n}}$ are in A.P and we have been asked to find the value of the series $\dfrac{1}{\sqrt{{{a}_{1}}}+\sqrt{{{a}_{2}}}}+\dfrac{1}{\sqrt{{{a}_{2}}}+\sqrt{{{a}_{3}}}}+......\dfrac{1}{\sqrt{{{a}_{n-1}}}+\sqrt{{{a}_{n}}}}$. So, for doing this type of questions we will use the concept of A.P and its common difference. After using that we will simplify the solution further using basic operations like addition and solve the question.

Complete step by step solution:
Firstly, from the given question we will take the first term and simplify it by multiplication the numerator and denominator with same $\left( \sqrt{{{a}_{1}}}-\sqrt{{{a}_{2}}} \right)$ term. So, we get the expression reduced as follows.
$\Rightarrow \dfrac{1}{\left( \sqrt{{{a}_{1}}}+\sqrt{{{a}_{2}}} \right)}$
Now, we will multiply this term with $\left( \sqrt{{{a}_{1}}}-\sqrt{{{a}_{2}}} \right)$ term with both numerator and denominator. So, we get,
$\Rightarrow \dfrac{1}{\left( \sqrt{{{a}_{1}}}+\sqrt{{{a}_{2}}} \right)}\times \dfrac{\left( \sqrt{{{a}_{1}}}-\sqrt{{{a}_{2}}} \right)}{\left( \sqrt{{{a}_{1}}}-\sqrt{{{a}_{2}}} \right)}$
$\Rightarrow \dfrac{\left( \sqrt{{{a}_{1}}}-\sqrt{{{a}_{2}}} \right)}{\left( {{a}_{1}}-{{a}_{2}} \right)}$
From the definition of A.P we know that the common difference between two successive terms is $d$. So, the expression can be further simplified as follows.
$\Rightarrow \dfrac{\left( \sqrt{{{a}_{1}}}-\sqrt{{{a}_{2}}} \right)}{d}$
We got the simplified first term of the series as follows.
$\Rightarrow \dfrac{\left( \sqrt{{{a}_{1}}}-\sqrt{{{a}_{2}}} \right)}{d}$
In the similar way the second and term and the next successive terms can be written in the reduced form as follows.
The second term can be written as follows.
$\Rightarrow \dfrac{\left( \sqrt{{{a}_{2}}}-\sqrt{{{a}_{3}}} \right)}{d}$
The third term of the series can be written as follows.
$\Rightarrow \dfrac{\left( \sqrt{{{a}_{3}}}-\sqrt{{{a}_{4}}} \right)}{d}$
So, the given question $\dfrac{1}{\sqrt{{{a}_{1}}}+\sqrt{{{a}_{2}}}}+\dfrac{1}{\sqrt{{{a}_{2}}}+\sqrt{{{a}_{3}}}}+......\dfrac{1}{\sqrt{{{a}_{n-1}}}+\sqrt{{{a}_{n}}}}$ can be rewritten in a simplified manner as follows.
$\Rightarrow \dfrac{1}{\sqrt{{{a}_{1}}}+\sqrt{{{a}_{2}}}}+\dfrac{1}{\sqrt{{{a}_{2}}}+\sqrt{{{a}_{3}}}}+......\dfrac{1}{\sqrt{{{a}_{n-1}}}+\sqrt{{{a}_{n}}}}$
$\Rightarrow \dfrac{\sqrt{{{a}_{1}}}-\sqrt{{{a}_{2}}}}{d}+\dfrac{\sqrt{{{a}_{2}}}-\sqrt{{{a}_{3}}}}{d}+......\dfrac{\sqrt{{{a}_{n-1}}}-\sqrt{{{a}_{n}}}}{d}$
$\Rightarrow \dfrac{\sqrt{{{a}_{1}}}-\sqrt{{{a}_{2}}}+\sqrt{{{a}_{2}}}-\sqrt{{{a}_{3}}}+.......\sqrt{{{a}_{n-1}}}-\sqrt{{{a}_{n}}}}{d}$
$\Rightarrow \dfrac{\sqrt{{{a}_{1}}}-\sqrt{{{a}_{n}}}}{d}$
Now, we will further simplify our solution by multiplying with $\sqrt{{{a}_{1}}}+\sqrt{{{a}_{n}}}$ on both numerator and denominator. So, we get,
$\Rightarrow \dfrac{\sqrt{{{a}_{1}}}-\sqrt{{{a}_{n}}}}{d}\times \dfrac{\sqrt{{{a}_{1}}}+\sqrt{{{a}_{n}}}}{\sqrt{{{a}_{1}}}+\sqrt{{{a}_{n}}}}$
$\Rightarrow \dfrac{{{a}_{1}}-{{a}_{n}}}{d\left( \sqrt{{{a}_{1}}}+\sqrt{{{a}_{n}}} \right)}$
Now, we can also write the expression with common difference as follows.
$\Rightarrow \dfrac{d\left( n-1 \right)}{d\left( \sqrt{{{a}_{1}}}+\sqrt{{{a}_{n}}} \right)}$
$\Rightarrow \dfrac{\left( n-1 \right)}{\left( \sqrt{{{a}_{1}}}+\sqrt{{{a}_{n}}} \right)}$
Therefore, the solution will be $\Rightarrow \dfrac{\left( n-1 \right)}{\left( \sqrt{{{a}_{1}}}+\sqrt{{{a}_{n}}} \right)}$.

Note: Students must be very careful in doing the calculations. Students must have good knowledge in the concept of A.P. students must not do mistakes like in the term if we use $\dfrac{d\left( n \right)}{d\left( \sqrt{{{a}_{1}}}+\sqrt{{{a}_{n}}} \right)}$ instead of $\Rightarrow \dfrac{d\left( n-1 \right)}{d\left( \sqrt{{{a}_{1}}}+\sqrt{{{a}_{n}}} \right)}$ our whole solution will be wrong.