Question

Three positive integers a1,a2,a3 are in AP such that a1+a2+a3=33 and a1*a2*a3=1155.Find the integers a1,a2,a3.

Answer 1

Given 
three  positive integers a1, a2,a3, in AP
a1*a2*a3*=1155\


=> a1+a2+a3=33..............................eq1
according to AP
2[a2]=a1+a3
2[a2]=33-a2    [from 1]
3[a2]=33
a2=11----------------------eq2

a1*a2*a3=1155
a1*a3=1155/11
a1*a3=105..........eq3

from eq 1 
a1+a2+a3=33
a1+a3=22
=>   a1=22-a3

a1*a3=105    
[22-a3] a3=105
22a3-[a3]square=105
[a3]square -22a3 +105=0

by factorizing we get 
a3=15 or 7............eq 4 

we have a3  and a2
 we can find  rremaining a1  
we get a1=7 or 15 

Answer 2

Answer:

Step-by-step explanation:

Let us consider the three positive integer in AP are

a k d 1 = - , a k 2 = , a k d 3 = +

Given, a a a 1 2 3 + + = 33

Þ k d k k d - + + + = 33

Þ 3 33 k = Þ k = 11 (1)

and a a a 1 2 3 ´ ´ = 1155

Þ( ) ( ) 11 11 11 1155 - ´ ´ + = d d

Þ ( )( ) 11 11 1155

11

- + = d d

(1)

Þ 121 105 2

- = d

\ d

2

= Þ 16 d = ± 4

\ k = 11, d = 4 or k = 11, d = - 4

When k = 11 and d = 4

a k d 1 = - = - = 11 4 7

a k d 2 = = = 11,

a k d 3 = + = + = 11 4 15,

When k = 11and d = - 4 (1)

or a k d 1 = - = + = 11 4 15

a k 2 = = 11 or a k d 3 = + = - = 11 4 7

Therefore, the numbers are 7, 11, 15 or 15, 11, 7.