Question

Sum of three numbers in AP is 12 and the sum of their squares is 56 find the numbers

Answer 1

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★--------Here is your answer ---★

Let The Number be a - d , a , a + d

So ----------->

a - d + a + a + d = 12
3a = 12
a = 4

So a = 4

Now ----->

( a - d )2 + a2 + (a + d)2 = 56
a2 + d2 - 2ad +a2+ a2 +d2 +2ad = 56
3a2 + 2d2 = 56
3× (4)2 + 2d2 = 56
48 + 2d2 = 56
2d2 = 8
d2 = 4
d = 2

So d = 2

Hence the number are

a - d = 4-2 = 2
a = 4
and a+d = 4+2=6
★
So , 2,4,6

Hope correct
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Answer 2

Solution:

Given: Sum of 3 terms of an AP is 12  and sum of their squares is 56.

To find: The numbers of that property

let a number be : a , then the others be: a+d, a-d

so,
 (a-d)+a+(a+d) =12
                     3a =12
                       a =4..(i)

(a-d)²+(a)²+(a+d)²=56
(4-d)²+(4)²+(4+d)²=56
(4)²-2(4)(d)+d²
+(4)²
+(4)²+2(4)(d)+(d)²=56

4²+d²+4²+4²+d²=56
        16+16+16+2d²=56
               48+2d²=56
                      2d²=56-48
                      2d²=8
                        d²=4
                         d=+2 (or) d=-2
hence, a+d=4-2=2 (or) a+d=4+2=6
     and, a-d=4-(-2)=4+2=6 (or) a-d=4-2=2

         Hope it Helps