Three numbers are in Arithmetic Progression (A.P.) whose

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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

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 =	30	= 10
	3

Again, a (a – d) (a + d) = 910
⇒ 10 (10 – d) (10 + d) = 910
⇒ 100 – d² = 91
⇒ d² = 100 – 91 = 9
⇒ d = √9 = 3
∴ Largest number = a + d = 10 + 3 = 13