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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Question : The sum of three positive numbers is 18 and their product is 162. If the sum of two numbers is equal to the third number, then the sum of the squares of the numbers is:
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