Question

) The sum of 4 consecutive terms of an AP
is 32 and the ratra of the pirniduct of the
first and last term to the product of
the middle term is 7:15 find the
numbers.​

Answer 1

Given Question:-

• The sum of 4 consecutive terms of an AP is 32 and the ratio of the product of the first and last term to the product ofthe middle term is 7:15. Find the numbers.

─━─━─━─━─━─━─━─━─━─━─━─━─

$\tt \: \huge \orange{AηsωeR} ✍$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\begin{gathered}\begin{gathered}\bf Given - \begin{cases} &\tt{4 \: numbers \: are \: in \: AP} \\ &\tt{Sum \: of \: 4 \: numbers \: is \: 32} \\ \\ & \tt{\dfrac{Product \: of \: extremes}{Product \: of \: means} = \dfrac{7}{15} }\\ \end{cases}\end{gathered}\end{gathered}$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\begin{gathered}\begin{gathered}\bf To \: Find :- \begin{cases} &\tt{4 \: numbers \:} \end{cases}\end{gathered}\end{gathered}$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\large\underline{\bold{Solution :- }}$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\tt \: Since \: 4 \: numbers \: are \: in \: AP$

$\begin{gathered}\begin{gathered}\bf Let \: the \: numbers \: be - \begin{cases} &\sf{x - 3y} \\ &\sf{x - y}\\ &\sf{x + y}\\ &\sf{x + 3y} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\bf\red{1. \: According \: to \: statement}\end{gathered}$

$\tt \: Sum \: of \: 4 \: numbers \: is \: 32$

$\tt \: \implies \: x - \cancel{3y} + x - \cancel y + x + \cancel y + x + \cancel{3y} = 32$

$\tt \: \implies \: 4x = 32$

$\tt \: \implies \: x = \dfrac{32}{4}$

$\tt \: \implies \: x \: = \: 8$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\begin{gathered}\tt\red{2. \: According \: to \: statement \: }\end{gathered}$

$\tt \: \tt{\dfrac{Product \: of \: extremes}{Product \: of \: means} = \dfrac{7}{15} }$

$\tt \: \implies \: \dfrac{(x - 3y)(x + 3y)}{(x - y)(x + y)} = \dfrac{7}{15}$

$\tt \: \implies \: \dfrac{ {x}^{2} - {9y}^{2} }{ {x}^{2} - {y}^{2} } = \dfrac{7}{15}$

$\tt \: \implies \: {15x}^{2} - {135y}^{2} = 7 {x}^{2} - {7y}^{2}$

$\tt \: \implies \: {8x}^{2} = 128 {y}^{2}$

$\tt \: \implies \: {x}^{2} = {16y}^{2}$

☆ On substituting x = 8, we get

$\tt \: \implies \: {16y}^{2} = {8}^{2}$

$\tt \: \implies \: {16y}^{2} = 64$

$\tt \: \implies \: {y}^{2} = 4$

$\tt \: \implies \: y \: = \pm \: 2$

─━─━─━─━─━─━─━─━─━─━─━─━─

✏ Now, two cases arises.

☆ Case :- 1.

$\tt \: When \: x \: = \: 8 \: and \: y \: = \: 2$

$\begin{gathered}\begin{gathered}\bf \: 4 \: numbers \: are - \begin{cases} &\sf{x - 3y = 8 - 6 = 2} \\ &\sf{x - y} = 8 - 2 = 6\\ &\sf{x + y} = 8 + 2 = 10\\ &\sf{x + 3y} = 8 + 6 = 14 \end{cases}\end{gathered}\end{gathered}$

─━─━─━─━─━─━─━─━─━─━─━─━─

☆ Case :- 2.

$\tt \: When \: x \: = \: 8 \: and \: y \: = \: - \: 2$

$\begin{gathered}\begin{gathered}\bf \: 4 \: numbers \: are - \begin{cases} &\sf{x - 3y = 8 + 6 = 14} \\ &\sf{x - y} = 8 + 2 = 10\\ &\sf{x + y} = 8 - 2 = 6\\ &\sf{x + 3y} = 8 - 6 = 2 \end{cases}\end{gathered}\end{gathered}$

─━─━─━─━─━─━─━─━─━─━─━─━─