Question

Day 6,

Five numbers are in A.P., whose sum is 25 and product is 2520. If one of these five numbers is −1/2 , then the greatest number amongst them is

[Answer-16]

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Answer 1

$\large\underline{\sf{Solution-}}$

Given that, 5 numbers are in AP.

Let assume that

$\red{\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:5 \: numbers \: be\begin{cases} &\sf{a - 2d} \\ &\sf{a - d}\\ &\sf{a}\\ &\sf{a + d}\\ &\sf{a + 2d} \end{cases}\end{gathered}\end{gathered}}$

According to statement,

Sum of 5 numbers is 25.

$\rm :\longmapsto\:a - 2d + a - d + a + a + d + a + 2d = 25$

$\rm :\longmapsto\:5a = 25$

$\red{\rm \implies\:\boxed{ \tt{ \: a \: = \: 5 \: }} - - - (1)}$

Again, Given that,

Product of 5 numbers = 2520

$\rm :\longmapsto\:(a - 2d)(a - d)a(a + d)(a + 2d) = 2520$

On substituting the value of a, we get

$\rm :\longmapsto\:(5 - 2d)(5 - d)5(5 + d)(5 + 2d) = 2520$

$\rm :\longmapsto\:(5 - 2d)(5 - d)(5 + d)(5 + 2d) = 504$

$\rm :\longmapsto\:(5 - 2d)(5 + 2d)(5 - d)(5 + d) = 504$

$\rm :\longmapsto\:(25 - {4d}^{2} )(25 - {d}^{2} ) = 504$

$\rm :\longmapsto\:625 - 25 {d}^{2} - 100 {d}^{2} + {4d}^{4} = 504$

$\rm :\longmapsto\:625 - 125 {d}^{2} + {4d}^{4} = 504$

$\rm :\longmapsto\:625 - 125 {d}^{2} + {4d}^{4} - 504 = 0$

$\rm :\longmapsto\:121 - 125 {d}^{2} + {4d}^{4} = 0$

$\rm :\longmapsto\: {4d}^{4} - 125 {d}^{2} + 121 = 0$

$\rm :\longmapsto\: {4d}^{4} - 4{d}^{2} - 121 {d}^{2} + 121 = 0$

$\rm :\longmapsto\: {4d}^{2}( {d}^{2} - 1) - 121( {d}^{2} - 1) = 0$

$\rm :\longmapsto\: ({4d}^{2} - 121)( {d}^{2} - 1) = 0$

$\rm \implies\: {d}^{2} = \dfrac{121}{4} \: \: or \: \: {d}^{2} = 1$

$\rm \implies\:d = \pm \: \dfrac{11}{2} \: \: or \: \: d = \: \pm \: 1$

$\bf\implies \:d \: = \: 1, \: - 1, \: \dfrac{11}{2}, \: - \dfrac{11}{2}$

So, we have following cases :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf a & \bf d \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 5 & \sf 1 \\ \\ \sf 5 & \sf - 1 \\ \\ \sf 5 & \sf \dfrac{11}{2} \\ \\ \sf 5 & \sf - \dfrac{11}{2} \end{array}} \\ \end{gathered}$

Hence, 5 Numbers in AP are

$\red{\rm :\longmapsto\:When \: a = 5 \: and \: d = 1}$

$\red{\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:5 \: numbers \: be\begin{cases} &\sf{a - 2d = 5 - 2 = 3} \\ &\sf{a - d = 5 - 1 = 4}\\ &\sf{a = 5}\\ &\sf{a + d = 5 + 1 = 6}\\ &\sf{a + 2d = 5 + 2 = 7} \end{cases}\end{gathered}\end{gathered}}$

Case - 2

$\red{\rm :\longmapsto\:When \: a = 5 \: and \: d = - 1}$

$\red{\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:5 \: numbers \: be\begin{cases} &\sf{a - 2d = 5 + 2 = 7} \\ &\sf{a - d = 5 + 1 = 6}\\ &\sf{a = 5}\\ &\sf{a + d = 5 - 1 = 4}\\ &\sf{a + 2d = 5 - 2 = 3} \end{cases}\end{gathered}\end{gathered}}$

Case - 3

$\red{\rm :\longmapsto\:When \: a = 5 \: and \: d = 5.5}$

$\red{\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:5 \: numbers \: be\begin{cases} &\sf{a - 2d = 5 - 11 = - 6} \\ &\sf{a - d = 5 - 5.5 = - 0.5}\\ &\sf{a = 5}\\ &\sf{a + d = 5 + 5.5 = 10.5}\\ &\sf{a + 2d = 5 + 11 = 16} \end{cases}\end{gathered}\end{gathered}}$

Case - 4

$\red{\rm :\longmapsto\:When \: a = 5 \: and \: d \: = - \: 5.5}$

$\red{\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:5 \: numbers \: be\begin{cases} &\sf{a - 2d = 5 + 11 = 16} \\ &\sf{a - d = 5 + 5.5 = 10.5}\\ &\sf{a = 5}\\ &\sf{a + d = 5 - 5.5 = - 0.5}\\ &\sf{a + 2d = 5 - 11 = - 6} \end{cases}\end{gathered}\end{gathered}}$

So,

• If one of 5 number is - 1/2, then largest number is 16.