Question

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

Answer 1

Answer:

let numbers be a-d,a,a+d

3a = 21

a = 7

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

(7-d)(7+d) = 231/7 = 33

49-d² = 33

d² = 16

d=4 or -4

if d = 4: numbers are 3,7,11

if d= -4: numbers are 7,3,-1

but product of (3,7,11)=231

so numbers are 3,7,11

Answer 2

Answer:

⚠ _ Hello _⚠

Step-by-step explanation:

⚫ Solution ⤵

$let \: the \: three \: numbers \: be \: \\ (a - d)(a)(a + d) \\ according \: to \: question \\ sum \: of \: these \: no. \\ (a - d) + a + (a + d) = 21 \\ 3a = 21 \\ a = 7 \\ according \: to \: question \\ (7 - d)(7 + d)(7) = 231 \\ {7}^{2} - {d}^{2} \times 7 = 231 \\ 343 - 7 {d}^{2} = 231 \\ - 7 {d}^{2} = 231 - 343 \\ - 7 {d}^{2} = - 112 \\ {d}^{2} = 16 \\ d = 4 \\ put \: the \: value \: of \: a \: and \: d \: in \: the \: numbers \\ (7 - 4) = 3 \\ (7 + 4) = 11 \\ 7$

❇ Hope it's help