# Math Help - Help

1. ## Help

If a1,a2,...an are in A.P. where ai>0 then value of
1/(sq root a1+sq root a2) + 1/(sq root a2+sq root a3) +.......... 1/sq root an-1 + sq root an = ??

I did not knw how to write it in proper form but hope u understand my question

2. Originally Posted by cnpranav
If a1,a2,...an are in A.P. where ai>0 then value of
1/(sq root a1+sq root a2) + 1/(sq root a2+sq root a3) +.......... 1/sq root an-1 + sq root an = ??

I did not knw how to write it in proper form but hope u understand my question
Do you mean $\frac{1}{\sqrt{a1}+\sqrt{a2}}+ \frac{1}{\sqrt{a2}+ \sqrt{a3}}+ \cdot\cdot\cdot+ \frac{1}{\sqrt{a_{n-1}+ a_n}}$?
(Edited)

And a1, a2, ..., an are an arithmetic progression?

I would start by looking at simple examples.

3. Originally Posted by HallsofIvy
Do you mean $\frac{1}{\sqrt{a1}+\sqrt{a2}}+ \frac{1}{\sqrt{a2}+ \sqrt{a3}}+ \cdot\cdot\cdot+ \sqrt{a_{n-1}+ a_n}$?

And a1, a2, ..., an are an arithmetic progression?

I would start by looking at simple examples.
the last expression is also in upon i.e. $frac{1}/{\sqrt{an-1}+\sqrt{an}}$

4. Hello, cnpranav!

If $a_1,a_2,a_3\:\hdots\:a_n$ are in A.P. where $a_i>0$

then find the value of: . $S \:=\:\frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + \hdots + \frac{1}{\sqrt{a_{n-1}} + \sqrt{a_n}}$
Rationalize the fractions . . .

. . . . . $\frac{1}{\sqrt{a_2} + \sqrt{a_1}}\cdot\frac{\sqrt{a_2}-\sqrt{a_1}}{\sqrt{a_2} - \sqrt{a_1}} \:=\: \frac{\sqrt{a_2}-\sqrt{a_1}}{a_2-a_1}$

. . . . . $\frac{1}{\sqrt{a_3} + \sqrt{a_2}}\cdot\frac{\sqrt{a_3} - \sqrt{a_2}}{\sqrt{a_3} - \sqrt{a_2}} \:=\: \frac{\sqrt{a_3} - \sqrt{a_2}}{a_3-a_2}$

. . . . . . . . . . . $\vdots$ . . . . . . . . . . . . . $\vdots$

. . $\frac{1}{\sqrt{a_n} + \sqrt{a_{n-1}}}\cdot\frac{\sqrt{a_n} - \sqrt{a_{n-1}}}{\sqrt{a_n} - \sqrt{a_{n-1}}} \:=\:\frac{\sqrt{a_n} - \sqrt{a_{n-1}}}{a_n - a_{n-1}}$

Since $a_1,a_2,a_3,\hdots$ are in A.P.,
. . all the denominators are the common difference, $d.$

We have: . $S \;=\;\frac{a_2-a_1}{d} + \frac{a_3-a_2}{d} + \frac{a_4-a_3}{d} + \hdots + \frac{a_n - a_{n-1}}{d}$

. . which simplifies to: . $S \;=\;\frac{a_n - a_1}{d}$

5. Originally Posted by Soroban
Hello, cnpranav!

Rationalize the fractions . . .

. . . . . $\frac{1}{\sqrt{a_2} + \sqrt{a_1}}\cdot\frac{\sqrt{a_2}-\sqrt{a_1}}{\sqrt{a_2} - \sqrt{a_1}} \:=\: \frac{\sqrt{a_2}-\sqrt{a_1}}{a_2-a_1}$

. . . . . $\frac{1}{\sqrt{a_3} + \sqrt{a_2}}\cdot\frac{\sqrt{a_3} - \sqrt{a_2}}{\sqrt{a_3} - \sqrt{a_2}} \:=\: \frac{\sqrt{a_3} - \sqrt{a_2}}{a_3-a_2}$

. . . . . . . . . . . $\vdots$ . . . . . . . . . . . . . $\vdots$

. . $\frac{1}{\sqrt{a_n} + \sqrt{a_{n-1}}}\cdot\frac{\sqrt{a_n} - \sqrt{a_{n-1}}}{\sqrt{a_n} - \sqrt{a_{n-1}}} \:=\:\frac{\sqrt{a_n} - \sqrt{a_{n-1}}}{a_n - a_{n-1}}$

Since $a_1,a_2,a_3,\hdots$ are in A.P.,
. . all the denominators are the common difference, $d.$

We have: . $S \;=\;\frac{a_2-a_1}{d} + \frac{a_3-a_2}{d} + \frac{a_4-a_3}{d} + \hdots + \frac{a_n - a_{n-1}}{d}$

. . which simplifies to: . $S \;=\;\frac{a_n - a_1}{d}$

after rationalizing the numerator was in under root how did it change??

6. the answer is given in the form n-1/(sq root a1+ sq root an)