Question

Find three numbers in an A.P such that their sum is 27 and the product is 504

Answer 1

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

Let assume that three numbers in AP series be

$\begin{gathered}\begin{gathered}\bf\: \begin{cases} &\sf{a - d} \\ &\sf{a}\\ &\sf{a + d} \end{cases}\end{gathered}\end{gathered}$

According to statement

Sum of three numbers in AP series is 27

$\rm\implies \:a - d + a + a + d = 27$

$\rm\implies \:3a = 27$

$\rm\implies \:\boxed{\tt{ \: a \: = \: 9 \: }}$

Also, According to statement

Product of three numbers is 504

$\rm\implies \:(a - d)a(a + d) = 504$

$\rm\implies \:a( {a}^{2} - {d}^{2} ) = 504$

On substituting the value of a, we get

$\rm\implies \:9( {9}^{2} - {d}^{2} ) = 504$

$\rm \: 81 - {d}^{2} = 56$

$\rm \: - {d}^{2} = 56 - 81$

$\rm \: - {d}^{2} = - 25$

$\rm \: {d}^{2} = 25$

$\rm\implies \:\boxed{\tt{ \: d \: = \: \pm \: 5 \: }}$

So, two cases arises

Case :- 1

When a = 9 and d = 5

So, three numbers in AP series are

$\begin{gathered}\begin{gathered}\bf\: \begin{cases} &\sf{a - d = 9 - 5 = 4} \\ \\ &\sf{a = 9}\\ \\ &\sf{a + d = 9 + 5 = 14} \end{cases}\end{gathered}\end{gathered}$

Case :- 2

When a = 9 and d = - 5

So, three numbers in AP series are

$\begin{gathered}\begin{gathered}\bf\: \begin{cases} &\sf{a - d = 9 + 5 = 14} \\ \\ &\sf{a = 9}\\ \\ &\sf{a + d = 9 - 5 = 4} \end{cases}\end{gathered}\end{gathered}$

VERIFICATION

Numbers in AP series are 4, 9, 14

So, their sum is 4 + 9 + 14 = 27

And

Product of these numbers = 4 × 9 × 14 = 504

Hence, Verified

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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.

↝ Sum of n  terms of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

Sₙ is the sum of n terms of AP.

a is the first term of the sequence.

n is the no. of terms.

d is the common difference.

Answer 2

Answer:

14, 9, 4 or 4, 9, 14

Step-by-step explanation:

let, first three teams

a+d, a, a-d

A/Q,

a+d+a+a-d = 27

3a = 27, a = 9.

and

(a+d).a.(a-d) = 504

(9+d).9.(9-d) = 504

9(81-d²) = 504

81-d² = 56

d² = 81-56 = 25

d = ±5

Now,

A.P first three teams , when d = 5

14, 9, 4

A.P first three teams , when d = -5

4, 9, 14