Question

find three numbers in ap whose sum is 15 and product is 105

Answer 1

Hey mate..
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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

Hope it helps !!

Answer 2

hello Vignesh ✋

⬇ here is your answer

Given as

Sum of three terms = 15 Product = 105

Since, we don't know the terms, therefore we will let as

a - d , a and a + d

According to Question

a - d + a + a + d = 15
3a = 15

a = 15

Hence we got the second term a

Now,

>>(a - d) ( a ) ( a + d ) = 105

> > (a^2 - ad)(a + d) = 105

> > a^3 - a^2 d + a^2 d - a d^2= 105

> > {a}^{3} - a {d}^{2} = 105

putting a = 5

> > 125 - 5 {d}^{2} = 105
> > 20 = 5 {d}^{2}

d = 2 or d = - 2

So, We got a and d both,

Now Find the required terms of AP

When d =2

first term = a - d = 5 - 2 = 3
Second term = a = 5
Third term = a + d = 5 + 2 = 7

3, 5 and 7

When d = -2
First term = a - d = 5 - ( - 2) = 7
Second term = a = 5
Third term = a + d = 5 - 2 = 3

7, 5 and 3

hope it helps
jerri