SOLUTION: Find 5 consecutive terms of an arithmetic progression whose sum is 115 and sum of squares of its second and fourth term is 1156.
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Question 1014387: Find 5 consecutive terms of an arithmetic progression whose sum is 115 and sum of squares of its second and fourth term is 1156. Answer by ikleyn(52858) (Show Source):
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Find 5 consecutive terms of an arithmetic progression whose sum is 115 and sum of squares of its second and fourth term is 1156.
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