Question

find 3 numbers in AP where sum is 21 and their product is 231

Answer 1

Let the numbers be a - d, a and a +d

Sum = 21

=> a - d + a + a + d = 21

=> 3a = 21

=> a = 7

Now,

Product = 231

=> (a-d) (a) (a + d) = 231

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

=> (7)^2 - (d)^2 = 33

=> 49 - d^2 = 33

=> d^2 = 49-33

=> d^2 = 16

=> d = 4

Required numbers are 3, 7, 11

Answer 2

Let number are ( a - d ), a, ( a + d )



$\textcolor{red}{given \: \: \: \: their \: \: sum \: = 21}$


=> a - d + a + a + d = 21

=> 3a = 21

=> a = 7



$\textcolor{blue}{their \: \: product = 231}$

=> ( a - d ) a ( a + d ) = 231

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

=> 343 - 7d² = 231

=> 343 - 231 = 7d²

=> 112 = 7d²

=> 16 = d²

=> 4 or - 4 = d




Hence,

Numbers are :

( I ) : ( 7 - 4 ) = 3 or ( 7 + 4 ) = 11

( II ): 7

( iii ): ( 7 + 4 ) = 11 or ( 7 - 4 ) = 3




$\fcolorbox{red}{lime}{numbers \: are \: (3 \: 7 \: and \: 11) \: \: or \: \: (11 \: \: 7 \: \: \: 3)}$