JEE 2026| geometric progression | Just check easy question | easy & tricky question | By Hiten sir скачать в хорошем качестве

JEE 2026| geometric progression | Just check easy question | easy & tricky question | By Hiten sir

JEE 2026| geometric progression | Just check easy question | easy & tricky question | By Hiten sir The video explains how to solve a math problem related to geometric progression (GP) for the JEE 2026 exam. Here's a breakdown of the key steps and concepts discussed: Understanding the problem: The problem involves an increasing GP (0:21), meaning the common ratio 'r' is greater than 1 (0:28). Setting up equations: The product of the second, third, and fourth terms (A2, A3, A4) is given as 64 (0:41). This leads to the equation ar³ = 64, from which ar = 4 (1:09). The sum of the first, third, and fifth terms (A1, A3, A5) is given as 813/7 (1:20). This translates to a + ar² + ar⁴ = 813/7 (1:20-1:33). Solving for 'a' and 'r': The instructor demonstrates how to substitute 'a' in terms of 'r' into the second equation (4:06), leading to a quadratic equation in terms of t (where t = r²) (4:34-5:38). The quadratic equation is 28t² - 785t + 28 = 0 (5:24-5:38). Finding r²: The video then shows how to factor the quadratic equation (5:56), resulting in t = r² = 28 (7:03-7:14). The other solution is rejected because 'r' must be greater than 1 for an increasing GP (7:23-7:27). Final Solution: With r² = 28, the final expression requested in the problem, A3 + A5 + A7 which is ar² + ar⁴ + ar⁶ simplifies to 4(1 + r² + r⁴) (2:35-3:12). Substituting the value of r² gives the final answer (7:34-7:42).