Question:
Five numbers are in A.P. whose sum is 25 and product is 2520. If one of these five numbers is -\( {1 \over 2} \) , then the greatest number amongst them is:
Five numbers are in A.P. whose sum is 25 and product is 2520. If one of these five numbers is -\( {1 \over 2} \) , then the greatest number amongst them is:
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Take the five terms as a-2d, a-d, a, a+d, a+2d; the sum fixes the middle term, and the product gives an equation in d^2. Then pick the case that contains -1/2.
Updated On: Oct 31, 2026
- \( {{21} \over 2} \)
- 27
- 7
- 16
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The Correct Option is D
Solution and Explanation
Let the numbers be a-2d, a-d, a, a+d, a+2d. Sum = 5a = 25, so a = 5.
Product: \(5(25-d^2)(25-4d^2)=2520\Rightarrow (25-d^2)(25-4d^2)=504\).
With \(u=d^2\): \(4u^2-125u+121=0\Rightarrow u=\dfrac{125\pm117}{8}\Rightarrow u=1\) or \(u=\dfrac{121}{4}\).
If \(d^2=1\), the numbers are 3, 4, 5, 6, 7 (no -1/2). So \(d^2=\frac{121}{4}\), \(d=\pm\frac{11}{2}\).
The numbers are 16, 10.5, 5, -0.5, -6 (or the same in reverse). Check: -0.5 is present, sum = 25.
Greatest number = 16.
Answer: (D) 16
Product: \(5(25-d^2)(25-4d^2)=2520\Rightarrow (25-d^2)(25-4d^2)=504\).
With \(u=d^2\): \(4u^2-125u+121=0\Rightarrow u=\dfrac{125\pm117}{8}\Rightarrow u=1\) or \(u=\dfrac{121}{4}\).
If \(d^2=1\), the numbers are 3, 4, 5, 6, 7 (no -1/2). So \(d^2=\frac{121}{4}\), \(d=\pm\frac{11}{2}\).
The numbers are 16, 10.5, 5, -0.5, -6 (or the same in reverse). Check: -0.5 is present, sum = 25.
Greatest number = 16.
Answer: (D) 16
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