# Arithmetic Progression Problem...

• March 23rd 2009, 01:37 AM
siddscool19
Arithmetic Progression Problem...
If $a_{1},a_{2},a_{3}.........a_{n}$ are in AP,where $a_{1}$>0, then the value of the expression
$\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt {a_{2}}+\sqrt{a_{3}}}+\frac{1}{\sqrt{a_{3}}+\sqrt{ a_{4}}}+........$ up to n terms:

Here are the options:

1. $\frac{1-n}{\sqrt{a_{1}}+\sqrt{a_{n}}}$

2. $\frac{n-1}{\sqrt{a_{1}}+\sqrt{a_{n}}}$

3. $\frac{n-1}{\sqrt{a_{1}}-\sqrt{a_{n}}}$

4. $\frac{1-n}{\sqrt{a_{1}}-\sqrt{a_{n}}}$
• March 23rd 2009, 02:36 AM
chisigma
If the sequence $a_{k}$ is an arithmetic progression then…

$a_{k}= a_{1} + (k-1)\cdot a$

Now we arrive to compute the sum in such a way…

$S= \sum_{k=2}^{n} \frac{1}{\sqrt{a_{k}}+\sqrt {a_{k-1}}}= \sum_{k=2}^{n} \frac{\sqrt{a_{k}} - \sqrt {a_{k-1}}}{a_{k} - a_{k-1}}=$

$= \frac {1}{a}\cdot (\sqrt{a_{2}} -\sqrt{a_{1}} + \sqrt{a_{3}} - \sqrt{a_{2}} + \dots + \sqrt {a_{n}} - \sqrt{a_{n-1}}) =$

$= \frac {1}{a}\cdot (\sqrt {a_{n}} - \sqrt{a_{1}}) = \frac {1}{a}\cdot (\sqrt {a_{1} + (n-1)\cdot a} - \sqrt{a_{1}})$

Kind regards

$\chi$ $\sigma$
• March 23rd 2009, 02:58 AM
n0083
Quote:

Originally Posted by siddscool19
If $a_{1},a_{2},a_{3}.........a_{n}$ are in AP,where $a_{1}$>0, then the value of the expression
$\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt {a_{2}}+\sqrt{a_{3}}}+\frac{1}{\sqrt{a_{3}}+\sqrt{ a_{4}}}+........$ up to n terms:

Here are the options:

1. $\frac{1-n}{\sqrt{a_{1}}+\sqrt{a_{n}}}$

2. $\frac{n-1}{\sqrt{a_{1}}+\sqrt{a_{n}}}$

3. $\frac{n-1}{\sqrt{a_{1}}-\sqrt{a_{n}}}$

4. $\frac{1-n}{\sqrt{a_{1}}-\sqrt{a_{n}}}$

Throughout we use
$a_n = a_1 + (n-1)d$ (*)

$\frac{1}{\sqrt{a_{k}}+\sqrt{a_{k+1}}} = \frac{\sqrt{a_{k}}-\sqrt{a_{k+1}}}{(\sqrt{a_{k}}+\sqrt{a_{k+1}})(\sqr t{a_{k}}-\sqrt{a_{k+1}})}$
$=\frac{\sqrt{a_{k}}-\sqrt{a_{k+1}}}{a_k-a_{k+1}}=\frac{\sqrt{a_{k}}-\sqrt{a_{k+1}}}{a_1+(k-1)d+(a_1-kd)}=\frac{\sqrt{a_{k}}-\sqrt{a_{k+1}}}{-d}$

Now each term has a common denominator namely $-d$. Adding each term up we see the middle terms in the numerator cancel leaving,
$\frac{\sqrt{a_{1}}-\sqrt{a_{n}}}{-d}$ (**)

using (*) again we have $d = \frac{a_n-a_1}{n-1}$.
Substituting this into (**) we have,
$\frac{(\sqrt{a_{1}}-\sqrt{a_{n}})(1-n)}{a_n-a_1}$
Factoring the denominator we have,
$\frac{(\sqrt{a_{1}}-\sqrt{a_{n}})(1-n)}{(\sqrt{a_n}-\sqrt{a_1})(\sqrt{a_n}+\sqrt{a_1})}$
Leaving,
$\frac{n-1}{\sqrt{a_n}+\sqrt{a_1}}$
which is answer choice 2 (thanks chisigma)
• March 23rd 2009, 03:20 AM
chisigma
Quote:

Originally Posted by chisigma
If the sequence $a_{k}$ is an arithmetic progression then…

$a_{k}= a_{1} + (k-1)\cdot a$

Now we arrive to compute the sum in such a way…

$S= \sum_{k=2}^{n} \frac{1}{\sqrt{a_{k}}+\sqrt {a_{k-1}}}= \sum_{k=2}^{n} \frac{\sqrt{a_{k}} - \sqrt {a_{k-1}}}{a_{k} - a_{k-1}}=$

$= \frac {1}{a}\cdot (\sqrt{a_{2}} -\sqrt{a_{1}} + \sqrt{a_{3}} - \sqrt{a_{2}} + \dots + \sqrt {a_{n}} - \sqrt{a_{n-1}}) =$

$= \frac {1}{a}\cdot (\sqrt {a_{n}} - \sqrt{a_{1}}) = \frac {1}{a}\cdot (\sqrt {a_{1} + (n-1)\cdot a} - \sqrt{a_{1}})$

In order to give a complete answer we consider that...

$(\sqrt{a_{n}}- \sqrt{a_{1}}) = \frac{a_{n} - a_{1}}{\sqrt{a_{n}} + \sqrt{a_{1}}}= \frac{(n - 1)\cdot a}{\sqrt{a_{n}} + \sqrt{a_{1}}}$

... so that is...

$S= \frac{n - 1}{\sqrt{a_{n}} + \sqrt{a_{1}}}$

The correct choice is 2)...

Kind regards

$\chi$ $\sigma$
• March 23rd 2009, 03:59 AM
Soroban
Hello, siddscool19!

Quote:

If $a_1,a_2,a_3, \hdots a_n$ are in AP, where $a_1 > 0$, find the value of the expression:

. . $S \;=\;\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{ \sqrt{a_{2}}+\sqrt{a_{3}}}+\frac{1}{\sqrt{a_{3}}+\ sqrt{a_{4}}}+ \hdots + \frac{1}{\sqrt{a_n} + \sqrt{a_{n+1}}}$

Here are the options:

$(1)\;\frac{1-n}{\sqrt{a_{1}}+\sqrt{a_{n}}} \qquad (2)\;\frac{n-1}{\sqrt{a_{1}}+\sqrt{a_{n}}} \qquad (3)\;\frac{n-1}{\sqrt{a_{1}}-\sqrt{a_{n}}} \qquad (4)\; \frac{1-n}{\sqrt{a_{1}}-\sqrt{a_{n}}}$

I don't agree with any of their answers.
Did I miscount?

Let $d$ be the common difference of the AP.

The $k^{th}$ term is: . $\frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}}$

Multiply top and bottom by: $\sqrt{a_{k+1}} - \sqrt{a_k}$

. . $\frac{1}{\sqrt{a_k} + \sqrt{a_{k+1}}} \cdot\frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{\sqrt{a_{k+1}} - \sqrt{a_k}} \;=\;\frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{\underbrace{a_{k+1} - a_k}_{\text{This is }d}} \;=\;\frac{\sqrt{a_{k+1}} - \sqrt{a_k}}{d}$

The series becomes:

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

Most of the terms cancel out and we are left with: . $S \;=\;\frac{\sqrt{a_{n+1}} - \sqrt{a_1}}{d}$

$\text{Rationalize: }\;S \;=\;\frac{\sqrt{a_{n+1}} - \sqrt{a_1}}{d} \cdot\frac{\sqrt{a_{n+1}} + \sqrt{a_1}}{\sqrt{a_{n+1}} + \sqrt{a_1}} \;=\; \frac{a_{n+1} - a_1}{d\left(\sqrt{a_{n+1}} + \sqrt{a_1}\right)}$

This can be simplified further, with our knowledge of AP's.

We know that: . $a_{n+1} \:=\:a_1 + nd$

So the numerator is: . $(a_1 + nd) - a_1 \:=\:nd$

And the fraction becomes: . $S \;=\;\frac{nd}{d\left(\sqrt{a_{n+1}} + \sqrt{a_1}\right)}$

Therefore: . $S \;=\;\frac{n}{\sqrt{a_{n+1}} + \sqrt{a_1}}$

. . This is closest to their answer-choice (2).

• March 23rd 2009, 04:34 AM
siddscool19