अनुक्रम \(4,7,12,19,28,\ldots\) में \(a_n=n^2+3\) है। कौन सा पद (84) है?

In the sequence \(4,7,12,19,28,\ldots\), \(a_n=n^2+3\). Which term is (84)?

Author: Muft Shiksha Editorial Team Published:
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Correct Answer

B. (9)वाँ(9)th

Step 1

Concept

From \(n^2+3=84\), \(n^2=81\) and (n=9). Exam tip: recognizing perfect squares saves time.

Step 2

Why this answer is correct

The correct answer is B. (9)वाँ / (9)th. From \(n^2+3=84\), \(n^2=81\) and (n=9). Exam tip: recognizing perfect squares saves time.

Step 3

Exam Tip

\(n^2+3=84\) से \(n^2=81\) और (n=9) मिलता है। वर्ग संख्या पहचानना समय बचाता है।

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अनुक्रम \(4,7,12,19,28,\ldots\) में \(a_n=n^2+3\) है। कौन सा पद (84) है? / In the sequence \(4,7,12,19,28,\ldots\), \(a_n=n^2+3\). Which term is (84)?

Correct Answer: B. (9)वाँ / (9)th. Explanation: \(n^2+3=84\) से \(n^2=81\) और (n=9) मिलता है। वर्ग संख्या पहचानना समय बचाता है। / From \(n^2+3=84\), \(n^2=81\) and (n=9). Exam tip: recognizing perfect squares saves time.

Which concept should I revise for this Mathematics MCQ?

From \(n^2+3=84\), \(n^2=81\) and (n=9). Exam tip: recognizing perfect squares saves time.

What exam hint can help solve this Mathematics question?

\(n^2+3=84\) से \(n^2=81\) और (n=9) मिलता है। वर्ग संख्या पहचानना समय बचाता है।