Let the first term α and the common ratio r of a geometric progression be positive integers. If the sum of squares of its first three terms is 33033, then the sum of these three terms is equal to

Solution

Maths sir · 4 steps

Factor 33033 completely. Since a² must divide it, only one first term is possible.

1. Set up. The terms are a, ar, ar², so the sum of their squares is a²(1 + r² + r⁴) = 33033.
2. Factor the number. 33033 = 3 × 7 × 11² × 13. The only perfect squares that divide it are 1 and 121, so a = 1 or a = 11.
3. Test a = 11. Then 1 + r² + r⁴ = 33033/121 = 273, so r⁴ + r² - 272 = 0. This factors as (r² - 16)(r² + 17) = 0, which gives r² = 16 and r = 4. With a = 1 you would need r⁴ + r² = 33032. That fails, because r = 13 gives 28730 and r = 14 gives 38612.

   a + ar + ar² = 11 + 44 + 176 = 231

4. So. The three terms are 11, 44 and 176. Their squares add to 121 + 1936 + 30976 = 33033, and their sum is 231, option (b).

The idea to keep. When integer unknowns hide inside a product like a² × (something), factor the given number into primes first. The square factor can take only a few values, so test each one.

Answer from the NTA final answer key. Solution by PrepWiser.