Question

If the sum of three numbers in an AP is 9 and their product is 24, then numbers are *
2 points
2, 4, 6
1, 3 , 5
1, 3, 8
2, 3, 4​

Answer 1

Answer:

2 3 4

Step-by-step explanation:

1. suppose tge three no. are (a+1) , (a), (a-1)

2. according to given condition ,

(a+1)+a+(a-1)=9

3a=9

a=9/3

a=3

3. put the value in given equation

a+1=3+1=4

a=3

a-1=3-1=2

so the numbers are 2,3,4

Answer 2

EXPLANATION.

$\sf : \implies \: let \: the \: three \: number \: are \: in \: ap \\ \sf : \implies \: (a - d) ,a ,(a + d)$

$\sf : \implies \: sum \: of \: three \: number \: in \: ap \: is \: = 9 \\ \\ \sf : \implies \: a - d + a + a + d \: = 9 \\ \\ \sf : \implies \: 3a \: = 9 \\ \\ \sf : \implies \: a \: = 3$

$\sf : \implies \: product \: \: is \: \: 24 \\ \\ \sf : \implies \: (a - d)(a)(a + d) = 24 \\ \\ \sf : \implies \: ( {a}^{2} - {d}^{2} )(a) = 24 \\ \\ \sf : \implies \: ( {3}^{2} - {d}^{2})(3) = 24 \\ \\ \sf : \implies \: (9 - {d}^{2} )(3) = 24$

$\sf : \implies \: 27 - 3 {d}^{2} = 24 \\ \\ \sf : \implies \: - 3 {d}^{2} = - 3 \\ \\ \sf : \implies \: d \: = \pm \: 1$

Therefore,

=> First term = a = 3

=> Common difference = d = ± 1

Case = 1.

=> ( a - d ) , a, ( a + d)

=> ( 3 - 1 ) , 3 , ( 3 + 1 )

=> 2,3,4

Case = 2.

=> ( a - d) , a, ( a + d)

=> ( 3 - (-1) ) , 3 , ( 3 + (-1))

=> 4,3,2

ADDITIONAL INFORMATION.

Nth term of an Ap.

Case = 1.

=> An = a + ( n - 1 ) d

=> An = 3 + ( n - 1 ) 1

=> An = 3 + n - 1

=> An = n + 2

Therefore,

=> If n = 1 => 1 + 2 = 3

=> if n = 2 => 2 + 2 = 4

=> if n = 3 => 3 + 2 = 5

=> if n = 4 => 4 + 2 = 6

Series are = 3,4,5,6,......

Case = 2.

Nth term of an Ap.

=> An = a + ( n - 1 ) d

=> An = 3 + ( n - 1 ) -1

=> An = 3 - n + 1

=> An = 4 - n

Therefore,

=> If n = 1 => 4 - 1 = 3

=> if n = 2 => 4 - 2 = 2

=> if n = 3 => 4 - 3 = 1

=> if n = 4 => 4 - 4 = 0

Series are = 3,2,1,0,.....