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How many initial terms of the arithmetic progression (81,75,69,\ldots) have sum (315)?

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Answer and explanation

Correct answer: (7)

Solving (\frac{n}{2}[162-6(n-1)]=315) gives (n=7). Exam tip: the same sum formula works for decreasing progressions too.

Related tags

ApDecreasing-ApExpert

Frequently asked questions

What is the correct answer to this question?

(7)

Why is this the correct answer?

Solving (\frac{n}{2}[162-6(n-1)]=315) gives (n=7). Exam tip: the same sum formula works for decreasing progressions too.

Which subject and chapter does this question cover?

This is a Class 10 Mathematics question. Chapter: Arithmetic Progressions (AP). Topic: Finding the sum of the first $n$ terms of an AP.

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