Question

three terms are in A.p. whose sum is -3and product of their cubes is 512​

Answer 1

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

Since, it is given that three numbers are in AP,

So,

$\begin{gathered}\begin{gathered}\bf\: Let-\begin{cases} &\sf{first \: number = a - d} \\ &\sf{second \: number = a}\\ &\sf{third \: number = a + d} \end{cases}\end{gathered}\end{gathered}$

$\red{\rm :\longmapsto\:According \: to \: statement}$

Sum of three numbers = - 3.

$\rm :\longmapsto\:a - \cancel d + a + a + \cancel d = - 3$

$\rm :\longmapsto\:3a = - 3$

$\bf\implies \:a = - 1 - - - (1)$

Again,

$\red{\rm :\longmapsto\:According \: to \: statement}$

Product of the cubes of three numbers = 512

$\rm :\longmapsto\: {(a - d)}^{3} \times {a}^{3} \times {(a + d)}^{3} = 512$

$\rm :\longmapsto\: {(a - d)}^{3} \times {a}^{3} \times {(a + d)}^{3} = 8 \times 8 \times 8$

$\rm :\longmapsto\: {(a - d)}^{3} \times {a}^{3} \times {(a + d)}^{3} = {8}^{3}$

$\rm :\implies\:(a - d)a(a + d) = 8$

$\rm :\longmapsto\:a( {a}^{2} - {d}^{2}) = 8$

$\rm :\longmapsto\: - 1( {( - 1)}^{2} - {d}^{2}) = 8$

$\rm :\longmapsto\: 1 - {d}^{2} = - 8$

$\rm :\longmapsto\: {d}^{2} = 9$

$\bf\implies \:d \: = \: \pm \: 3$

Hence,

Two cases arises

Case :- 1

When a = - 1 and d = 3

So numbers are

$\begin{gathered}\begin{gathered}\bf\: Hence-\begin{cases} &\sf{first \: number = - 1 - 3 = - 4} \\ &\sf{second \: number = - 1}\\ &\sf{third \: number = - 1 + 3 = 2} \end{cases}\end{gathered}\end{gathered}$

Case :- 2

When a = - 1 and d = - 3

So, numbers are

$\begin{gathered}\begin{gathered}\bf\: Hence-\begin{cases} &\sf{first \: number = - 1 + 3 = 2} \\ &\sf{second \: number = - 1}\\ &\sf{third \: number = - 1 - 3 = - 4} \end{cases}\end{gathered}\end{gathered}$

Additional Information :-

↝ nᵗʰ term of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

• aₙ is the nᵗʰ term.

• a is the first term of the sequence.

• n is the no. of terms.

• d is the common difference.