अनुक्रम \(6,9,14,21,\ldots\) का (n)वाँ पद कौन-सा है?

What is the (n)th term of the sequence \(6,9,14,21,\ldots\)?

Author: Muft Shiksha Editorial Team Published:
Explanation opens after your attempt
Correct Answer

A. \(n^2+5\)

Step 1

Concept

The terms are \(1^2+5,2^2+5,3^2+5,\ldots\), so \(a_n=n^2+5\). Identify the square plus a constant.

Step 2

Why this answer is correct

The correct answer is A. \(n^2+5\). The terms are \(1^2+5,2^2+5,3^2+5,\ldots\), so \(a_n=n^2+5\). Identify the square plus a constant.

Step 3

Exam Tip

पद \(1^2+5,2^2+5,3^2+5,\ldots\) हैं, इसलिए \(a_n=n^2+5\)। वर्ग के साथ स्थिर जोड़ पहचानें।

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Mathematics Answer, Explanation and Revision Hints

अनुक्रम \(6,9,14,21,\ldots\) का (n)वाँ पद कौन-सा है? / What is the (n)th term of the sequence \(6,9,14,21,\ldots\)?

Correct Answer: A. \(n^2+5\). Explanation: पद \(1^2+5,2^2+5,3^2+5,\ldots\) हैं, इसलिए \(a_n=n^2+5\)। वर्ग के साथ स्थिर जोड़ पहचानें। / The terms are \(1^2+5,2^2+5,3^2+5,\ldots\), so \(a_n=n^2+5\). Identify the square plus a constant.

Which concept should I revise for this Mathematics MCQ?

The terms are \(1^2+5,2^2+5,3^2+5,\ldots\), so \(a_n=n^2+5\). Identify the square plus a constant.

What exam hint can help solve this Mathematics question?

पद \(1^2+5,2^2+5,3^2+5,\ldots\) हैं, इसलिए \(a_n=n^2+5\)। वर्ग के साथ स्थिर जोड़ पहचानें।