Question

THE SUM OF THE THREE CONSECUTIVE NUMBERS IN A.P. IS 21 AND THEIR PRODUCT IS 231. FIND THE NUMBERS SOLUTION:

Answer 1

Let the three consecutive terms be a, a+d, a+2d
Sum of the terms = 21
⇒ a+a+d+a+2d = 21
3a + 3d = 21
a+d = 7 ⇒ d = 7-a
Product of numbers = 231
⇒ a(a+d)(a+2d) = 231
    a(7)(a+2[7-a]) = 231
7a(a+14-2a) = 231
7a(14-a) = 231
98a - 7a² = 231
7a² - 98a + 231 = 0
a² - 14a + 33 = 0
a² - 11a - 3a +33 = 0
a(a-11) -3(a-11) = 0
(a-3)(a-11) = 0
a=3 or a = 11
If a=3 then d=4
the terms are 3,7,11
If a = 11 then d = -4
the terms are 11,7,3

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