Question

find the four numbers in an AP whose sum is 50 and in which the greatest number is 4 times the least

Answer 1

☆Hey friend!!!!☆

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Here is your answer ☞
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$let \: the \: four \: number \: in \: ap \: be \: \\ \\ a - 3d. \: \: a - d. \: \: a + d. \: \: a + 3d \\ \\ now \: according \: to \: question \: \\ \\ (a - 3d) + (a - d) + (a + d) + (a + 3d) \: = 50 \\ \\ = > 4a \: = 50 \\ \\ = > \: a \: = \frac{25}{2} \\ \\ greatest \: number \: = 4 \times least \: number \\ \\ = > \: (a + 3d) = 4 \times (a - 3d) \\ \\ = > \: ( \frac{25}{2} + 3d) = 4 \times ( \frac{25}{2} - 3d) \\ \\ = > \frac{25}{2} + 3d \: = 50 - 12d \\ \\ = > \: 15d \: = 50 - \frac{25}{2} \\ \\ = > 15d \: = \frac{75}{2} \\ \\ = > d = \frac{5}{2} \\ \\ then \: the \: number \: are \\ \\ = > a - 3d = \frac{25}{2} - 3 \times \frac{5}{2} \\ = > \: a - 3d \: = \: 5........eq _{1} \\ \\ = > a - d = \frac{25}{2} - \frac{5}{2} \\ = > \: a - d = 10 \: ..........eq _{2} \\ \\ = > a + d = \frac{25}{2} + \frac{5}{2} \\ = > \: a + d = 15 \: .......eq _{3} \\ \\ = > a + 3d \: = \frac{25}{2} + 3 \times \frac{5}{2} \\ = > a + 3d \: = 20 \: .......eq _{4}$

Hence, Number is AP are {5,10,15,20}

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I think my answer is capable to clear your confusion... ☺☺☺☺☺☺☺
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Devil_king ▄︻̷̿┻̿═━一