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Question
The next term of the A.P. \[\sqrt{7}, \sqrt{28}, \sqrt{63}\] is ______.
Options
- \[\sqrt{70}\]
`sqrt(84)`
- \[\sqrt{97}\]
- \[\sqrt{112}\]
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Solution
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
\[i.e. a = d = \sqrt{7}\]
The next term of the sequence, i.e. the 4th term = a + 3d
\[= \sqrt{7} + 3\sqrt{7}\]
\[ = 4\sqrt{7}\]
\[ = \sqrt{7 \times 4 \times 4}\]
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