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Question

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


Solution

Let the 4 numbers are a , a + d , a + 2d , a + 3d.

Sum of 4 numbers AP = 50

a + a + d + a + 2d + a + 3d = 50

⇒ 4a + 6d = 50

⇒ 2a + 3d = 25 --------------(1)

Also given the greatest number is 4 times the least.

4(a) = a + 3d

4a - a = 3d

a = d

putting a = d in (1) , we obtain

5d = 25

d = 5

a = 5 but d = a.

∴ First four terms are 5 , 10 ,15 , 20.


Mathematics
RD Sharma
Standard X

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