Question

If $a_1,a_2,a_3,....a_n$ are in A.P., where $a_1>0$ for all $i$, then $\frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+....+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}$ is equal to

• A. $\frac{1}{\sqrt{a_1}+\sqrt{a_n}}$
• B. $\frac{1}{\sqrt{a_1}-\sqrt{a_n}}$
• C. $\frac{n}{\sqrt{a_1}-\sqrt{a_n}}$
• D. $\frac{n-1}{\sqrt{a_1}+\sqrt{a_n}}$

Answer 1

The correct option is D $\frac{n-1}{\sqrt{a_1}+\sqrt{a_n}}$

$a_1,a_2,a_3........A.P.,So{a}_2-a_1=a_3-a_2=a_4-a_3=d$.....(i)

Now,

given series:-

$\frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+\frac{1}{\sqrt{a_3}+\sqrt{a_4}}+---------+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}$

$=\frac{1}{\sqrt{a_1}+\sqrt{a_2}}\times \frac{\sqrt{a_2}-\sqrt{a_1}}{\sqrt{a_2}-\sqrt{a_1}}+\frac{\sqrt{a_3}-\sqrt{a_2}}{\sqrt{a_3}+\sqrt{a_2}\times (\sqrt{a_3}-\sqrt{a_2})}+--------+\frac{\sqrt{a_{n-1}}-\sqrt{a_n}}{\sqrt{a_{n-1}}+\sqrt{a_n}\times (\sqrt{a_{n-1}}-\sqrt{a_n})}$

$=\frac{\sqrt{a_2}-\sqrt{a_1}}{a_2-a_1}+\frac{\sqrt{a_3}-\sqrt{a_2}}{a_3-a_2}+------\frac{\sqrt{a_n}-\sqrt{a_{n-1}}}{\sqrt{a_n}-\sqrt{a_{n-1}}}$

$=\frac{1}{d}[\sqrt{a_2}-\sqrt{a_1}+\sqrt{a_3}-\sqrt{a_2}+-------+\sqrt{a_n}-\sqrt{a_{n-1}}]$........................from (i)

$=\frac{1}{d}[\sqrt{a_2}-\sqrt{a_1}+\sqrt{a_3}-\sqrt{a_2}+\sqrt{a_4}-\sqrt{a_3}+-----+\sqrt{a_{n-1}}-\sqrt{a_{n-2}}+\sqrt{a_n}-\sqrt{a_{n+1}}]$

$=[\sqrt{a_n}-\sqrt{a_1}]\times \frac{1}{d}$................(ii)

now,

$a_n=a_1+(n-1)d$

$a_n-a_1=(n-1)d\Rightarrow d=\frac{a_n-a_1}{n-1}$.......(iii)

replacing it in (ii)

$=\frac{n-1}{(a_n-a_1)}\times (\sqrt{a_n}-\sqrt{a_1})=\frac{n-1}{(\sqrt{a_n}+\sqrt{a_1})}$.........(D)