Login/Sign Up
Home » If are in A.P., where for all i, show that
If a_{1}, a_{2}, a_{3}, \ldots, a_{n} are in A.P., where a_{i}\lt{0} for all i, show that \frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\cdots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_{n}}}=\frac{n-1}{\sqrt{a_{1}}+\sqrt{a_{n}}}
CBSE | Class 11 | Maths | NCERT Exemplar | Sequences and Series
If a_{1}, a_{2}, a_{3}, \ldots, a_{n} are in A.P., where a_{i}\lt{0} for all i, show that \frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\cdots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_{n}}}=\frac{n-1}{\sqrt{a_{1}}+\sqrt{a_{n}}}

Solution:

It is given that a_{1}, a_{2}, a_{3}, \ldots, a_{n} are in the form of AP in which a_{i}\lt{0} for all i.

To prove that:

\begin{array}{l} \frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\cdots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_{n}}}=\frac{n-1}{\sqrt{a_{1}}+\sqrt{a_{n}}} \\ \Rightarrow \text { LHS }=\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\cdots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_{n}}} \end{array}

Multiply first term by \frac{\sqrt{2_{1}}-\sqrt{a_{2}}}{\sqrt{a_{1}}-\sqrt{a_{2}}}, the second term by \frac{\sqrt{a_{2}}-\sqrt{a_{2}}}{\sqrt{a_{2}}-\sqrt{a_{3}}} and so on i.e., rationalizing each term

\Rightarrow Left Hand Side =\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}} \times \frac{\sqrt{a_{1}}-\sqrt{a_{2}}}{\sqrt{a_{1}}-\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}} \times \frac{\sqrt{a_{2}}-\sqrt{a_{3}}}{\sqrt{a_{2}}-\sqrt{a_{3}}}+\cdots +\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_{n}}} \times \frac{\sqrt{a_{n-1}}-\sqrt{a_{n}}}{\sqrt{a_{n-1}}-\sqrt{a_{n}}}

By using (a+b)(a-b)=a^{2}-b^{2}

\Rightarrow Left Hand Side =\frac{\sqrt{\mathrm{a}_{1}}-\sqrt{\mathrm{a}_{2}}}{\mathrm{a}_{1}-\mathrm{a}_{2}}+\frac{\sqrt{\mathrm{a}_{2}}-\sqrt{\mathrm{a}_{3}}}{\mathrm{a}_{2}-\mathrm{a}_{3}}+\cdots+\frac{\sqrt{\mathrm{a}_{\mathrm{n}-1}}-\sqrt{\mathrm{a}_{\mathrm{n}}}}{\mathrm{a}_{\mathrm{n}-1}-\mathrm{a}_{\mathrm{n}}}

Since a_{1}, a_{2}, a_{3}, \ldots, a_{n} are in AP let us assume ‘d’ as its common difference

a_{2}-a_{1}=d, a_{3}-a_{2}=d \ldots a_{n}-a_{n-1}=d

As a result multiplying by -1

a_{1}-a_{2}=-d, a_{2}-a_{3}=-d \ldots a_{n}-a_{n-1}=-d

On putting these values in the Left Hand Side, we get

\begin{array}{l} \Rightarrow \text{Left Hand Side}=\frac{\sqrt{\mathrm{a}_{1}}-\sqrt{\mathrm{a}_{2}}}{-\mathrm{d}}+\frac{\sqrt{\mathrm{a}_{2}}-\sqrt{\mathrm{a}_{3}}}{-\mathrm{d}}+\cdots+\frac{\sqrt{\mathrm{a}_{\mathrm{n}-1}}-\sqrt{\mathrm{a}_{\mathrm{n}}}}{\mathrm{a}_{\mathrm{n}-1}-\mathrm{a}_{\mathrm{n}}} \\ =-\frac{1}{\mathrm{~d}}\left(\sqrt{\mathrm{a}_{1}}-\sqrt{\mathrm{a}_{2}}+\sqrt{\mathrm{a}_{2}}-\sqrt{\mathrm{a}_{3}}+\cdots+\sqrt{\mathrm{a}_{\mathrm{n}-1}}-\sqrt{\mathrm{a}_{\mathrm{n}}}\right) \\ =-\frac{1}{\mathrm{~d}}\left(\sqrt{\mathrm{a}_{1}}-\sqrt{\mathrm{a}_{\mathrm{n}}}\right) \end{array}

In the above equation multiply and divide by

\left(\sqrt{a_{1}}+\sqrt{a_{n}}\right)

\begin{array}{l} \Rightarrow \text{Left Hand Side}=-\frac{1}{\mathrm{~d}} \frac{\left(\sqrt{\mathrm{a}_{1}}-\sqrt{\mathrm{a}_{\mathrm{n}}}\right)\left(\sqrt{\mathrm{a}_{1}}+\sqrt{\mathrm{a}_{\mathrm{n}}}\right)}{\left(\sqrt{\mathrm{a}_{1}}+\sqrt{\mathrm{a}_{\mathrm{n}}}\right)} \\ \text { Now using }(\mathrm{a}+\mathrm{b})(\mathrm{a}-\mathrm{b})=\mathrm{a}^{2}-\mathrm{b}^{2} \\ \Rightarrow \mathrm{Left Hand Side}=-\frac{1}{\mathrm{~d}} \frac{\mathrm{a}_{1}-\mathrm{a}_{\mathrm{n}}}{\left(\sqrt{\mathrm{a}_{1}}+\sqrt{\mathrm{a}_{\mathrm{n}}}\right.} \end{array}

The n^{\text {th }} term of the AP is t_{n}=a+(n-1) d

In which the t_{n}=a_{n} is last n^{\text {th }} term and
a=a_{1} is first term

As a result a_{n}=a_{1}+(n-1) d

\Rightarrow \mathrm{a}_{1}-\mathrm{a}_{\mathrm{n}}=-(\mathrm{n}-1) \mathrm{d}

Now, substitute \mathrm{a}_{1}-\mathrm{a}_{\mathrm{n}} in Left Hand Side

\begin{array}{l} \Rightarrow \mathrm{Left Hand Side}=-\frac{1}{\mathrm{~d}} \frac{-(\mathrm{n}-1) \mathrm{d}}{\left(\sqrt{\mathrm{a}_{1}}+\sqrt{\mathrm{a}_{\mathrm{n}}}\right)} \\ \left.=\frac{(\mathrm{n}-1)}{\left(\sqrt{\mathrm{a}_{1}}+\sqrt{\mathrm{a}_{\mathrm{n}}}\right.}\right) \end{array}

\Rightarrow {L.H.S.}={R.H.S.}

As a result hence proved.

i

More Material