Question

The sum of the three consecutive numbers in A.P.is 21 and their product is 231.Find the numbers.

Answer 1

hello Frnd

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

$( {a}^{2} - ad)(a + d) = 231 \\ {a}^{3} + {a}^{2} d - {a}^{2} d - a {d}^{2} = 231$
Now putting a = 7

${7}^{3} - 7 {d}^{2} = 231 \\ 343 - 7 {d}^{2} = 231 \\ 343 - 231 = 7 {d}^{2} \\ 112 \div 7 = {d}^{2} \\ d = \sqrt{16} \\ d = 4 \: \: \: \: or \: \: \: - 4$
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

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