Question

The next term of the A.P. \[\sqrt{7}, \sqrt{28}, \sqrt{63}\] is ______.

विकल्प

• \[\sqrt{70}\]

• \[\sqrt{84}\]

• \[\sqrt{97}\]
• \[\sqrt{112}\]

Answer 1

The next term of the A.P. \[\sqrt{7}, \sqrt{28}, \sqrt{63}\] is `underlinebb(sqrt(112))`.

Explanation:

We have:

\[\sqrt{7}, \sqrt{28}, \sqrt{63}\]

\[or \sqrt{7}, \sqrt{7 \times 2 \times 2}, \sqrt{7 \times 3 \times 3}\]

\[or \sqrt{7 \times 1 \times 1}, \sqrt{7 \times 2 \times 2}, \sqrt{7 \times 3 \times 3} . . .\]

\[or \sqrt{7}, 2\sqrt{7}, 3\sqrt{7} . . .\]

We can see that the first term of the given A.P. is

\[\sqrt{7}\] and the common difference is \[2\sqrt{7} - \sqrt{7}\] = \[\sqrt{7}\]

\[i.e. a = d = \sqrt{7}\]

The next term of the sequence, i.e. the 4^th term = a + 3d

\[= \sqrt{7} + 3\sqrt{7}\]

\[ = 4\sqrt{7}\]

\[ = \sqrt{7 \times 4 \times 4}\]

\[ = \sqrt{112}\]