Question

The sum of three consecutive number in an ap is 21 and their product 231 find the number

Answer 1

let a be first term and d be common difference
let no. in ap be a-d , a , a+d
A.T.Q
a-d+a+a+d=21
a=7
and
(a-d)(a+d)a=231
(7(49-d^2))=231
49-d^2=33
d^2=16
d=+-4
no. are
11 , 7, 3 or 3,7,11
hope this helps:)

Answer 2

Hey mate..
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Let the numbers be a−d ,a and a+d

Clearly d is the common difference.

A/Q,

( a−d ) + a + ( a+d ) =21

=> 3a = 21

=> a = 7

Also,

a (a−d) (a+d)=231

=> 7 (72−d^2)=231

=> 72−d^2=33

=> d^2=16

=> d=±4

So, the numbers are 7, 4, (7+4) ...

 i.e, 3,7,11...

Or,

The numbers are 11,7,3.....

Hope it helps !!!!