यदि \(a_1=80\) और \(a_{n+1}=a_n-\frac{7n}{2}\) है तो \(a_5\) क्या होगा?

If \(a_1=80\) and \(a_{n+1}=a_n-\frac{7n}{2}\), what is \(a_5\)?

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

A. (45)

Step 1

Concept

The total subtraction is (\frac{7(1+2+3+4)}{2}=35), so \(a_5=45\). Exam tip: add fractional subtractions first.

Step 2

Why this answer is correct

The correct answer is A. (45). The total subtraction is (\frac{7(1+2+3+4)}{2}=35), so \(a_5=45\). Exam tip: add fractional subtractions first.

Step 3

Exam Tip

कुल घटाव (\frac{7(1+2+3+4)}{2}=35) है इसलिए \(a_5=45\) है। भिन्न घटावों को पहले जोड़ें।

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Mathematics Answer, Explanation and Revision Hints

यदि \(a_1=80\) और \(a_{n+1}=a_n-\frac{7n}{2}\) है तो \(a_5\) क्या होगा? / If \(a_1=80\) and \(a_{n+1}=a_n-\frac{7n}{2}\), what is \(a_5\)?

Correct Answer: A. (45). Explanation: कुल घटाव (\frac{7(1+2+3+4)}{2}=35) है इसलिए \(a_5=45\) है। भिन्न घटावों को पहले जोड़ें। / The total subtraction is (\frac{7(1+2+3+4)}{2}=35), so \(a_5=45\). Exam tip: add fractional subtractions first.

Which concept should I revise for this Mathematics MCQ?

The total subtraction is (\frac{7(1+2+3+4)}{2}=35), so \(a_5=45\). Exam tip: add fractional subtractions first.

What exam hint can help solve this Mathematics question?

कुल घटाव (\frac{7(1+2+3+4)}{2}=35) है इसलिए \(a_5=45\) है। भिन्न घटावों को पहले जोड़ें।