Question

The sequence $\frac{5}{\sqrt{7}},\frac{6}{\sqrt{7}},\sqrt{7}....$ is

• A. HP
• B. GP
• C. AP
• D. None of these

Answer 1

The correct option is C

AP

Explanation for the correct option:

Option(C)

An arithmetic progression (AP) is a sequence where the differences between every two consecutive terms are the same.

Let $\frac{5}{\sqrt{7}},\frac{6}{\sqrt{7}},\sqrt{7}....$ be represented as $a,b,c,....$ where $d$is difference between two consecutive terms.

Check the common difference.

$$\begin{gathered} b-a=\left(\frac{6}{\sqrt{7}}\right)-\left(\frac{5}{\sqrt{7}}\right)=\left(\frac{1}{\sqrt{7}}\right) \\ c-b=\left(\sqrt{7}\right)-\left(\frac{6}{\sqrt{7}}\right)=\left(\frac{1}{\sqrt{7}}\right) \\ d=\left(\frac{1}{\sqrt{7}}\right) \end{gathered}$$

Here $b-a=c-b$

So it is an AP.

Explanation for the in-correct option:

Option(A):

$\frac{5}{\sqrt{7}},\frac{6}{\sqrt{7}},\frac{7}{\sqrt{7}}...$ in HP

If $\frac{\sqrt{7}}{5},\frac{\sqrt{7}}{6},\frac{\sqrt{7}}{7}...$ are in AP

$2\frac{\sqrt{7}}{6}\ne \frac{\sqrt{7}}{5}+\frac{\sqrt{7}}{7}$

Hence option (A) is not correct.

Option(B)

$\frac{5}{\sqrt{7}},\frac{6}{\sqrt{7}},\frac{7}{\sqrt{7}}...$in GP

If ${\left(\frac{6}{\sqrt{7}}\right)}^{2}isequalto\frac{5}{\sqrt{7}}\times \frac{7}{\sqrt{7}}$

But it is not possible

Hence option(B) is not correct.

Hence the correct option is option $C$.