Question

the sum of the three consecutive numbers ina.p is 21 and their product is 231 find the numbers

Answer 1

hello Kunal
here is your solution

Let the three terms of AP be 

a - d , a and a + d 

According to question 

a - d + a + a + d = 21 

3a = 21 

a = 7 

Now, it's given that 

(a - d)( a) ( a + d ) = 231 
${a}^{3} + {a}^{2}d - {a}^{2} d - a {d}^{2} = 231$

Now putting a = 7 
${7}^{3} - 7 {d}^{2} = 231 \\ 343 - 231 = 7 {d}^{2} \\ d = \sqrt{16} \\ d = 4 \: \: or \: \: - 4$

Now the required AP is 

When a = 7 and d = 4 

3, 7 , 11 

when a = 7 and d = -4 

11, 7 , 3 

hope it helps
#jerri

Answer 2

Here is your solution

let,

Three numbers be a,a-d,a+d

so a+a-d+a+d = 21

a = 7

A/q
(a)(a-d)(a+d) = 231

(7)(7-d)(7+d) = 231

49- d2 = 231/7

d2 = 16

therefore d = +4 or - 4

so no(s) r : 3, 7 and 10 or 10, 7 and 3

hope it helps you