Sir\Madam please explain me this question on Arithmetic Progressions!!

Asked by  | 11th Jan, 2010, 02:41: PM

Expert Answer:

Let the four integers be, a-3d, a-d, a+d, and a+3d.

Their sum,

a-3d + a-d + a+d + a+3d = 4a = 24,

a = 6

and product,

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

(a2-9d2)(a2-d2) = 945

(36 - 9d2)(36 - d2) = 945

9(4 - d2)(36 - d2) = 945

(4 - d2)(36 - d2) = 105

d4 - 40d2 + 144 - 105 = 0

d4 - 40d2 + 39 = 0

Solving this quadratic equation in d2

d2 = {40±(1600-(4x39)}/2

= {40±38}/2

Hence d =1 and the four integers are,

3, 5, 7, 9

Regards,

Team,

TopperLearning.

 

 

Answered by  | 11th Jan, 2010, 03:02: PM

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