Question

Find sum of three numbers is an A.p in 27 and their product is 405.find the number

Answer 1

Heya!!Here is your answer friend ⤵⤵
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$let \: the \: terms \: be \: \: \: a - d \: \: \: a \: \: and \: a + d \\ now \: given \: that \: there \: sum = 27 \\ a - d + a + a + d = 27 \\ 3a = 27 \\ a = 9 \\ \\ hence \: terms \: become \: \: 9 - d \: \: \: \: 9 \: \: and \: 9 + d \\ now \: given \: there \: product = 405 \\ (9 - d)(9 + d) \times 9 = 405 \\ (81 - d {}^{2} ) \times 9 = 405 \\ 721 - 9d {}^{2} - 405 = 0 \\ 9d {}^{2} + 405 - 729= 0 \\ \\ 9d { }^{2} = 324 \\ d {}^{2} = 36 \\ d = \frac{ + }{ - } 6 \\ hence \: the \: terms \: ar \: e \\ \\ (1) \: taking \: + ve \: value(6) \: \\ = 3 \: \: 9 \: \: and \: 15 \\ \\ (2)taking \: - ve \: value \: ( - 6) \\ \\ the \: terms \: become \: \\ = 15 \: \: 9 \: and \: 3$

Hope it helps you ✌✌☺

Answer 2

pl mark it as brainliest