Let be a GP of increasing positive numbers. If the product of fourth and sixth terms is 9 and the sum of fifth and seventh terms is 24, then is equal to ________.
Show answer and solution
Answer: 60
Solution
Let the GP have first term a and common ratio r. The fourth term is ar³, sixth is ar⁵, fifth is ar⁴, seventh is ar⁶.
Product of 4th and 6th: ar³ · ar⁵ = a²r⁸ = 9, so ar⁴ = 3 (taking positive root since terms are positive). This means the 5th term a₅ = 3.
Sum of 5th and 7th: ar⁴ + ar⁶ = ar⁴(1 + r²) = 3(1 + r²) = 24, so 1 + r² = 8, r² = 7...
Wait, let me recheck: 3(1 + r²) = 24 → r² = 7. Then r = √7 and a = 3/r⁴ = 3/49.
Now compute each part:
- a₅ = 3, a₇ = ar⁶ = 3r² = 21, so a₅ + a₇ = 24 ✓
- a₁a₉: a₁a₉ = a · ar⁸ = a²r⁸ = 9
- a₂a₄a₉: a₂ = ar, a₄ = ar³, a₉ = ar⁸, so a₂a₄a₉ = a³r¹² = (ar⁴)³ = 3³ = 27
Therefore a₁a₉ + a₂a₄a
Product of 4th and 6th: ar³ · ar⁵ = a²r⁸ = 9, so ar⁴ = 3 (taking positive root since terms are positive). This means the 5th term a₅ = 3.
Sum of 5th and 7th: ar⁴ + ar⁶ = ar⁴(1 + r²) = 3(1 + r²) = 24, so 1 + r² = 8, r² = 7...
Wait, let me recheck: 3(1 + r²) = 24 → r² = 7. Then r = √7 and a = 3/r⁴ = 3/49.
Now compute each part:
- a₅ = 3, a₇ = ar⁶ = 3r² = 21, so a₅ + a₇ = 24 ✓
- a₁a₉: a₁a₉ = a · ar⁸ = a²r⁸ = 9
- a₂a₄a₉: a₂ = ar, a₄ = ar³, a₉ = ar⁸, so a₂a₄a₉ = a³r¹² = (ar⁴)³ = 3³ = 27
Therefore a₁a₉ + a₂a₄a
[JEE Main 2023 · 29 Jan · Shift 1]