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in Arithmetic Progression by (56.3k points)

Find the four numbers in A.P., whose sum is 50 and in which the greatest number is 4 times the least.

1 Answer

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Let’s consider the four terms of the A.P. to be (a – 3d), (a – d), (a + d) and (a + 3d).

From the question,

Sum of these terms = 50

⇒ (a – 3d) + (a – d) + (a + d) + (a + 3d) = 50

⇒ a – 3d + a – d + a + d + a – 3d= 50

⇒ 4a = 50

⇒ a = 50/4 = 25/2

And, also given that the greatest number = 4 x least number

⇒ a + 3d = 4 (a – 3d)

⇒ a + 3d = 4a – 12d

⇒ 4a – a = 3d + 12d

⇒3a = 15d

⇒a = 5d

Using the value of a in the above equation, we have

⇒25/2 = 5d

⇒ d = 5/2

So, the terms will be:

(a – 3d) = (25/2 – 3(5/2)), (a – d) = (25/2 – 5/2), (25/2 + 5/2) and (25/2 + 3(5/2)).

⇒ 5, 10, 15, 20

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