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