If the sum and product of four positive consecutive terms of a G.P., are 126 and 1296 , respectively, then the sum of common ratios of all such GPs is

Solution

Maths sir · 4 steps

Write the four terms symmetrically around the middle, and the product hands you the first term at once.

1. Symmetric terms. Take the four terms as a/r³, a/r, ar, ar³ with a, r > 0, so the common ratio is r². Their product is a⁴ = 1296 = 6⁴, so a = 6.
2. Use the sum. 6(r³ + 1/r³ + r + 1/r) = 126, so r³ + 1/r³ + r + 1/r = 21. Put t = r + 1/r, so that r³ + 1/r³ = t³ - 3t. This gives t³ - 2t - 21 = 0.

   t³ - 2t - 21 = (t - 3)(t² + 3t + 7) = 0

3. Only real root. The quadratic t² + 3t + 7 has discriminant 9 - 28 < 0, so it has no real roots. Hence r + 1/r = 3.
4. So. r + 1/r = 3 has two positive roots, r and 1/r. They give the two GPs with ratios r² and 1/r², which are the same terms read in reverse. The sum of the ratios is r² + 1/r² = 3² - 2 = 7, option (a).

The idea to keep. For four terms of a GP, use a/r³, a/r, ar, ar³ so the product is just a⁴. Then r + 1/r becomes the natural variable. Remember that the real common ratio here is r².

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