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Question
Product of the fourth term and the fifth term of an arithmetic progression is 456. Division of the ninth term by the fourth term of the progression gives Quotient as 11 and the remainder as 10. Find the first term of the progression.
Product of the fourth term and the fifth term of an arithmetic progression is 456. Division of the ninth term by the fourth term of the progression gives Quotient as 11 and the remainder as 10. Find the first term of the progression.
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Solution
The correct option is D - 66
A factor search for factor pairs of 456 gives us the following possibilities. 1,456; 2,228; 3,152; 4,114; 6,76; 8,57; 12,38 and 19,24. A check of the conditions given in the problem tells us that if we take 12 as the 4th term and 38 as the 5th term, we would get the series till 9 terms as - 66, - 40, - 14, 12, 38, 64, 90, 116, 142. In this series, we can see that the division of the 9th term by the 4th term gives us a quotient of 11 and a remainder of 10. Hence, the required first term is -66.
A factor search for factor pairs of 456 gives us the following possibilities. 1,456; 2,228; 3,152; 4,114; 6,76; 8,57; 12,38 and 19,24. A check of the conditions given in the problem tells us that if we take 12 as the 4th term and 38 as the 5th term, we would get the series till 9 terms as - 66, - 40, - 14, 12, 38, 64, 90, 116, 142. In this series, we can see that the division of the 9th term by the 4th term gives us a quotient of 11 and a remainder of 10. Hence, the required first term is -66.
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