Home › ICSE Class 10 › mathematics › Arithmetic Progression › How many terms of the AP 1, 4, 7, ... sum to 287?
How many terms of the AP 1, 4, 7, ... sum to 287?
A16
B12
C15
D14
Answer & Solution
Correct answer: D. 14
1. a = 1, d = 3 and S_n = n/2 * (2 + 3(n - 1)) = n(3n - 1)/2.
2. Set n(3n - 1)/2 = 287, so 3 n^2 - n - 574 = 0.
3. Factor: (3n + 41)(n - 14) = 0.
4. Take positive n = 14.
_Source: Selina Concise Mathematics Class 10, Ch on Arithmetic Progression (Mistral OCR'd PDF), section on sum_
Related questions
Rupees 720 was divided among A, B, C, D and E. The amounts received were in ascending ordeIf S_n = n^2 + 3 n, then first term and common difference areHow many three-digit numbers are divisible by 6?If sum of an AP is 100 and a = 5, l = 15, number of terms isIf a_n = 3 n + 5, the first three terms areSum of all natural numbers between 100 and 200 divisible by 7 isMiddle term of the AP 1, 5, 9, ..., 49 isSum of an AP with first term 7, last term 47 and 9 terms is