अनुक्रम \(4,11,22,37,56,\ldots\) का (n)वाँ पद \(a_n=n^2+3n+1\) है। (12)वाँ पद क्या होगा?

The sequence \(4,11,22,37,56,\ldots\) has (n)th term \(a_n=n^2+3n+1\). What will be the (12)th term?

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Correct Answer

B. (181)

Step 1

Concept

\(a_{12}=12^2+3\times12+1=181\). Exam tip: substitute the term position for (n).

Step 2

Why this answer is correct

The correct answer is B. (181). \(a_{12}=12^2+3\times12+1=181\). Exam tip: substitute the term position for (n).

Step 3

Exam Tip

\(a_{12}=12^2+3\times12+1=181\) है। सूत्र में (n) की जगह पद संख्या रखें।

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अनुक्रम \(4,11,22,37,56,\ldots\) का (n)वाँ पद \(a_n=n^2+3n+1\) है। (12)वाँ पद क्या होगा? / The sequence \(4,11,22,37,56,\ldots\) has (n)th term \(a_n=n^2+3n+1\). What will be the (12)th term?

Correct Answer: B. (181). Explanation: \(a_{12}=12^2+3\times12+1=181\) है। सूत्र में (n) की जगह पद संख्या रखें। / \(a_{12}=12^2+3\times12+1=181\). Exam tip: substitute the term position for (n).

Which concept should I revise for this Mathematics MCQ?

\(a_{12}=12^2+3\times12+1=181\). Exam tip: substitute the term position for (n).

What exam hint can help solve this Mathematics question?

\(a_{12}=12^2+3\times12+1=181\) है। सूत्र में (n) की जगह पद संख्या रखें।