हिंदी

The next term of the A.P. 7,28,63 is ______.

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प्रश्न

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

विकल्प

  • \[\sqrt{70}\]
  • `sqrt(84)`

  • \[\sqrt{97}\]
  • \[\sqrt{112}\]
MCQ
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उत्तर

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 4th term = a + 3d

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

\[ = 4\sqrt{7}\]

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

\[ = \sqrt{112}\]
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अध्याय 5: Arithmetic Progressions - Exercise 5.8 [पृष्ठ ६०]

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आर.डी. शर्मा Mathematics [English] Class 10
अध्याय 5 Arithmetic Progressions
Exercise 5.8 | Q 41 | पृष्ठ ६०

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