यदि \(a_1=120\) और (a_{n+1}=a_n-\(n!+2^n\)) है तो \(a_5\) क्या होगा?

If \(a_1=120\) and (a_{n+1}=a_n-\(n!+2^n\)), what is \(a_5\)?

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

D. (57)

Step 1

Concept

The subtracted values are (3,6,14,40), so \(a_5=57\). Exam tip: find the sum of all subtractions first.

Step 2

Why this answer is correct

The correct answer is D. (57). The subtracted values are (3,6,14,40), so \(a_5=57\). Exam tip: find the sum of all subtractions first.

Step 3

Exam Tip

घटने वाले मान (3,6,14,40) हैं इसलिए \(a_5=57\) है। पूरे घटावों का योग पहले निकालें।

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Mathematics Answer, Explanation and Revision Hints

यदि \(a_1=120\) और (a_{n+1}=a_n-\(n!+2^n\)) है तो \(a_5\) क्या होगा? / If \(a_1=120\) and (a_{n+1}=a_n-\(n!+2^n\)), what is \(a_5\)?

Correct Answer: D. (57). Explanation: घटने वाले मान (3,6,14,40) हैं इसलिए \(a_5=57\) है। पूरे घटावों का योग पहले निकालें। / The subtracted values are (3,6,14,40), so \(a_5=57\). Exam tip: find the sum of all subtractions first.

Which concept should I revise for this Mathematics MCQ?

The subtracted values are (3,6,14,40), so \(a_5=57\). Exam tip: find the sum of all subtractions first.

What exam hint can help solve this Mathematics question?

घटने वाले मान (3,6,14,40) हैं इसलिए \(a_5=57\) है। पूरे घटावों का योग पहले निकालें।