Question

Find three nos. in A.P whose sum is 21 and their product is 336

Answer 1

Let the numbers be a - d, a and a + d.  Sum = a - d + a + a + d = 3a = 21 => a = 3.  And also, (a-d)a(a+d) = 336  a(a2 - d2) = 336  7(49 - d2) = 336  343 - 7d2 = 336 => d2 = 1 => d = ±1  Hence numbers can be 6,7,8 or 8,7,6 depending on what you take as d. But as both give same set of numbers, the three numbers are 6, 7 and 8 (order is not important)

Answer 2

✌✌Here is your answer ✌ ✌

$\huge\boxed{\red{\bold{solution}}}$

$\large\underline{\purple{\bold{Given}}}$

☆sum of ap=21.
☆Their products of ap is 336.

Let ,
The Ap is (a-d) , a , (a+d)
A/q
(a-d)+a +(a+d)=21
a-d+a+a+d=21
3a=21
a=7✔

Their products is 336
so,
(a-d) × a × (a+d) =336
(7-d)×7× (7+d)= 336
(7-d)×(7+d) =336/7
49-d^2=48
49-48=d^2
1=d✔

So hence,

Ap are (a-d) , a , (a+d)

Ap are ➡(7-1) , 7, (7+1)

Ap are ➡6,7,8,.......

$\huge\underline{\purple{\bold{Thanks}}}$
I hope it helps you ✔