Question

the sum of three numbers in ap is 18 if the product of first and third number is five times the common difference find the numbers.​

Answer 1

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

Given that, sum of 3 numbers in AP is 18.

Let assume that three numbers in AP be a - d, a, a + d.

So, we have

$\rm \: a - d + a + a + d = 18$

$\rm \: 3a = 18$

$\rm\implies \:\boxed{ \rm{ \:a \: = \: 6 \: }} - - - (1)$

Further given that, the product of first and third number is five times the common difference.

So, we have

$\rm \: (a - d)(a + d) = 5d$

On substituting the value of a, from equation (1), we get

$\rm \: (6 - d)(6 + d) = 5d$

$\rm \: 36 - {d}^{2} = 5d$

$\rm \: {d}^{2} + 5d - 36 = 0$

$\rm \: {d}^{2} + 9d - 4d - 36 = 0$

$\rm \: d(d + 9) - 4(d + 9) = 0$

$\rm \: (d + 9)(d - 4) = 0$

$\rm\implies \:d = - 9 \: \: or \: \: d = 4$

So, two cases arises.

Case :- 1 When a = 6 and d = 4

So, numbers are 2, 6, 10

Case :- 2 When a = 6 and d = - 9

So, numbers are 15, 6, - 3

$\rule{190pt}{2pt}$

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

PROVIDED INFORMATION :-

the sum of three numbers in ap is 18 if the product of first and third number is five times the common difference

QUESTION :-

the sum of three numbers in ap is 18 if the product of first and third number is five times the common difference find the numbers.

TO FIND :-

find the numbers = ?

SOLUTION :-

Given sum of three numbers = 18

Product of 1 and 3 = 5 common difference

let numbers be (a - d), a, ( a+ d)

(a - d) + a + (a + d) = 18

3a = 18

a = 6

Also, (a - d)(a + d) = 5d

a ^ 2 - d ^ 2 = 5d

6 ^ 2 - d ^ 2 = 5d

d ^ 2 + 5d - 36 = 0

d ^ 2 + 9d - 4d - 36 = 0

d ( d +9) - 4 ( d + 9) = 0

(d - 4) (d + 9) = 0

d = 4, - 9

Numbers are 2, 6, 10 or 15,6, -3