In an AP, a = 5, and the sum from the 16th term to the 30th term is 1395. Find d.
Answer and explanation
Correct answer: 4
There are 30 − 16 + 1 = 15 terms from the 16th through the 30th term. The first of these is t₁₆ = a + 15d = 5 + 15d, and the last is t₃₀ = a + 29d = 5 + 29d. Using the sum formula for these 15 terms, 1395 = 15/2[(5 + 15d) + (5 + 29d)] = 15/2(10 + 44d) = 75 + 330d. Hence 330d = 1320, so d = 4. Therefore option C is correct. Subtracting S₁₅ from S₃₀ gives the same result; forgetting that both endpoints are included would incorrectly count only 14 terms.
Frequently asked questions
What is the correct answer to this question?
4
Why is this the correct answer?
There are 30 − 16 + 1 = 15 terms from the 16th through the 30th term. The first of these is t₁₆ = a + 15d = 5 + 15d, and the last is t₃₀ = a + 29d = 5 + 29d. Using the sum formula for these 15 terms, 1395 = 15/2[(5 + 15d) + (5 + 29d)] = 15/2(10 + 44d) = 75 + 330d. Hence 330d = 1320, so d = 4. Therefore option C is correct. Subtracting S₁₅ from S₃₀ gives the same result; forgetting that both endpoints are included would incorrectly count only 14 terms.
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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