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
here is your answer

Let the three terms of an AP is

a - d, a , a + d

Now, according to question

a - d + a + a + d = 21
3a = 21
a = 7

Now,

(a - d ) a (a + d) = 231
$( {a}^{2} - ad)(a + d) = 231 \\ {a}^{3} - {a}^{2} d + {a}^{2} d - a {d}^{2} = 231 \\ {a}^{3} - a {d}^{2} = 231 \\ \\ \\ \\ \\ now \: putting \: a = 7 \\ \\ 3 43 - 7 {d}^{2} = 231 \\ - 7 {d}^{2} = 231 - 343 \\ 112 = 7 {d}^{2} \\ {d}^{2} = 16 \\ d = 4 \: \: \: \: \: \: \: or \: d \: = - 4$

Now, required AP is

❇When a = 7 Nd d = 4

a - d = 7 - 4= 3
a = 7
a + d = 7 + 4 = 11

Required AP is 3, 7 and 11

❇When a = 7 Nd d = -4

a - d = 7 - ( -4 ) = 11
a = 7
a + d = 7 - 4 = 3

Required AP is 11, 7 and 3



hope it helps
jerri