Question

The sum of the first four numbers in an arithmetic progression is 48 and the sum of their squares is 756, what is the common difference? [quovantis18]

Answer 1

Answer:

3, 9, 15, 21

Step-by-step explanation:

Let (a - 3d), (a - d), (a+d) and (a+3d) be the nos

ATQ,

       a - 3d + a - d + a + d + a + 3d = 48

         ⇒ 4a = 48

                a = 12        

(a - 3d)² + (a - d)² + (a+d)² + (a+3d)² = 756

(12 - 3d)² + (12 - d)² + (12+d)² + (12+3d)² = 756

(12² + 9d² - 72d)+(12² + d² - 24d)+(12² + d² + 24d)+(12² + 9d² + 72d)= 756

4 * 144 + 20d² = 756

144 + 5d² = 189

5d² = 189 - 144 = 45

d² = 9

d = ± 3

Thus the nums are   3, 9, 15, 21

or

                                21, 15, 9, 3

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