Question

If 4 numbers are in arithmetic progression such that their sum is 50 and the greatest number is four times the least find the numbers​

Answer 1

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.

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Answer 2

Answer:

least number= 5

greatest number= 20

AP: 5, 10, 15, 20

Step-by-step explanation:

Sn= 50; n= 4

Given that, an = 4a

Sn = n/2(a+an)

=>50 = 2(a + 4a)

=>50 = 2 × 5a

=>50 = 10a

=>a = 5

=>an = 4a

=>an = 4 × 5

=>an = 20

a + (n-1)d = an

=> 5 + 3d = 20

=> d = 15/3

=> d = 5