Question

find three numbers in AP whose sum is 15 and whose product is 105​

Answer 1

Hey friend, Here is your answer-

Let the three numbers are a-b , a , a+b

From first condition ,

a-b+a+a+b = 15

3a = 15

a= 15/3

a= 5

From second condition,

(a-b) ×a×(a+b) = 105

a-b ( a^2 + ab) = 105

5-b (5^2 + 5b) = 105

5-b (25 + 5b) = 105

125 + 25b-25b-5b^2=105

-5b^2 +125 = 105

-5b^2 = 105 - 125

-5b^2 = -20

5b^2 = 20

b^2 = 20/5

b^2 = 4

b = root 4

b = 2

a-b= 5-2 = 3

a= 5

a+b= 5+2= 7

Therefore , the three numbers in AP are

3 , 5 , 7.

Hope it will help you

Answer 2

Answer:

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Let the three number in A.P be (a – d), a and (a + d) respectively.

Given, The three numbers in ap whose sum is 15

So..

(a – d ) + a + (a + d ) = 15

=> 3a = 15

=>  a = 5

Again,

The three numbers in ap whose product is 105.

So,

(a – d) a (a + d) = 105

a(a^2 – d^2) = 105

5(25 – d^2) = 105

25 – d^2 = 21

d^2 = 4

d = ± 2

When , d = 2

The three terms of the A.P  

= (5 – 2), 5 and (5 + 2)

= 3, 5 and 7

And when, d = – 2

The three terms of the A.P = (5 + 2) 5, and (5 – 2)

= 7, 5 and 3.  

So, The numbers are 3,5,7