Question : Three numbers are in Arithmetic Progression (A.P.) whose sum is 30 and the product is 910. Then the greatest number in the A.P. is:
Option 1: 17
Option 2: 15
Option 3: 13
Option 4: 10
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Correct Answer: 13
Solution :
Let three numbers in A.P. be a – d, a, and a + d, respectively.
According to the question, a – d + a + a + d = 30
⇒ 3a = 30
⇒ a = $\frac{30}{3}$ = 10
Again, a×(a – d)×(a + d) = 910
⇒ 10×(10 – d)×(10 + d) = 910
⇒ 100 – d
2
= 91
⇒ d
2
= 100 – 91 = 9
⇒ d = $\sqrt{9}$ = 3
$\therefore$ Largest number = a + d = 10 + 3 = 13
Hence, the correct answer is 13.
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