Question

If $a_r>0,rϵN$ and $a_1$, $a_2$, $a_3$,..., $a_{2n}$ are in AP then
$\frac{a_1+a_{2n}}{\sqrt{a_1}+\sqrt{a_2}}+\frac{a_2+a_{2n-1}}{\sqrt{a_2}+\sqrt{a_3}}+\frac{a_3+a_{2n-2}}{\sqrt{a_3}+\sqrt{a_4}}+...+\frac{a_n+a_{n+1}}{\sqrt{a_n}+\sqrt{a_{n+1}}}$
is equal to

• A. n-1
• B. $\frac{n(a_1+a_{2n})}{\sqrt{a_1}+\sqrt{a_{n+1}}}$
• C. $\frac{n-1}{\sqrt{a_1}+\sqrt{a_{n+1}}}$
• D. none of these

Answer 1

The correct option is B $\frac{n(a_1+a_{2n})}{\sqrt{a_1}+\sqrt{a_{n+1}}}$

Given $a_1,a_2,a_3......,a_{2n}$ are in A.P
We know that $S_n=\frac{n}{2}(a_1+a_{2n})=\frac{n}{2}(a_2+a_{2n-1})$ and so on
where $S_n$ is the sum of the given A.P
Hence each numerator term can be written as $\frac{2S_n}{n}$
Now let's multiply the numerator and denominator by the common difference ${}^{\prime }{d}^{\prime }$
So the sequence $=$
$\frac{1}{d}\times \{\frac{(a_1+a_{2n})d}{\sqrt{a_1}+\sqrt{a_2}}+\frac{(a_2+a_{2n-1})d}{\sqrt{a_2}+\sqrt{a_3}}+.....+\frac{(a_n+a_{n+})d}{\sqrt{a_n}+\sqrt{a_{n+1}}}\}=\frac{1}{d}\times \{\frac{(a_1+a_{2n})(a_2-a_1)}{\sqrt{a_1}+\sqrt{a_2}}+\frac{(a_2+a_{2n-1})(a_3-a_2)}{\sqrt{a_2}+\sqrt{a_3}}+.....+\frac{(a_n+a_{n+})(a_{n+1}-a_n)}{\sqrt{a_n}+\sqrt{a_{n+1}}}\}$
Now substitute the value of original numerator in the sequence we get:
$\frac{2S_n}{n}\times \frac{1}{d}\times \{\frac{(a_2-a_1)}{\sqrt{a_1}+\sqrt{a_2}}+\frac{(a_3-a_2)}{\sqrt{a_2}+\sqrt{a_3}}+........+\frac{(a_{n+1}-a_n)}{\sqrt{a_n}+\sqrt{a_{n+1}}}\}=\frac{2S_n}{n}\times \frac{1}{d}\times \{(\sqrt{a_2}-\sqrt{a_1})+(\sqrt{a_3}-\sqrt{a_2})+.......+(\sqrt{a_{n+1}}-\sqrt{a_n})\}$
There will be cancellation of terms.
Hence the sequence now becomes :
$\frac{2S_n}{n}\times \frac{1}{d}\times (\sqrt{a_{n+1}}-\sqrt{a_1})=(a_1+a_{2n})\times n\times \frac{(\sqrt{a_{n+1}}-\sqrt{a_1})}{nd}=n(a_1+a_{2n})\times \frac{(\sqrt{a_{n+1}}-\sqrt{a_1})}{(a_{n+1}-a_1)}$
Here we have written $nd=(a_{n+1}-a_1)$
Since $a_{n+1}=a_1+nd$
$=n(a_1+a_{2n})\times \frac{1}{(\sqrt{a_{n+1}}+\sqrt{a_1})}=\frac{n(a_1+a_{2n})}{\sqrt{a_{n+1}}+\sqrt{a_1}}$