n-th term, sum, AP word problems, sum-property identitiesPractice Questions & MCQs
Practice n-th term, sum, AP word problems, sum-property identities questions from Arithmetic Progressions (Class 10 Maths) - important MCQs with answers and step-by-step solutions. Solve the sample questions below, then practice the full set free on the RankByte app.
n-th term, sum, AP word problems, sum-property identitiesQuiz - Solve & Score
Q1. How many terms of the AP -10, -7, -4, ... are needed for sum 104?
- A.13
- B.11
- C.12
- D.14
Answer: A. 13
Given: -10; -7; -4; 104. Target quantity: How many terms of the AP -10, -7, -4. The relation that links these is - a = -10, d = 3. This is the equation that links the given quantities to the unknown (math, chapter 'Arithmetic Progressions'). Putting the numbers in: a = -10, d = 3 → S_n = n/2(-20 + 3(n-1)) = n(3n-23)/2 = 104 → 3n^2 - 23n - 208 = 0 → n = (23+√(529+2496))/6 = (23+55)/6 = 13. Hence the answer is A) 13.
Q2. How many three-digit numbers are divisible by 7?
- A.128
- B.127
- C.129
- D.130
Answer: A. 128
Start by listing the data - 7 (math, chapter 'Arithmetic Progressions'). What we must find: How many three-digit numbers are divisible by 7?. This is a classic application of Count = (994-105)/7 + 1 = 128. The arithmetic is: Count = (994-105)/7 + 1 = 128. First 3-digit multiple of 7: 105. Matching this against the options, A) 128 is the answer.
Q3. How many terms of the AP 9, 17, 25, ... must be taken to get sum 636?
- A.12
- B.14
- C.10
- D.13
Answer: A. 12
We are told that 9, 17, 25, 636 (math, chapter 'Arithmetic Progressions'). Our target: How many terms of the AP 9, 17, 25. From the chapter we use the relation: S_n = n/2(2·9 + (n-1)·8) = n(4n+5) = 636 → 4n^2+5n-636 = 0 → n = (-5+√(25+10176))/8 = (-5+101)/8 = 12. Crunching it out: S_n = n/2(2·9 + (n-1)·8) = n(4n+5) = 636 → 4n^2+5n-636 = 0 → n = (-5+√(25+10176))/8 = (-5+101)/8 = 12. So the correct choice is A) 12.
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