Question

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

Answer 1

Answer:

The four numbers in A.P are 5, 10, 15 and 20.

Step-by-step explanation:

Given :  

Four numbers in A.P., whose sum is 50

Let the four numbers in AP are  (a – 3d), (a – d), (a + d) and (a + 3d) and let d > 0 . Here smallest number is (a - 3d) and the greatest number is (a + 3d)

A.T.Q

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

4a = 50.

a = 25/2  ...............(1)

 

A.T.Q

(a + 3d) = 4(a – 3d)

a + 3d = 4a – 12d

a - 4a = - 12d - 3d

-3a = - 15d

3a = 15d

a = 15d/3

a = 5d  

5d = 25/2  

[From eq 1]

d = 25/2 × ⅕  

d = 5/2  

Therefore,

First number , (a – 3d) =  25/2 - 3(5/2) = 25/2 - 15/2 = (25 -15)/2 = 10/2 = 5

 Second number , (a – d) = 25/2 - 5/2 = 25/2 - 5/2 = (25 -5)/2 = 20/2 = 10

 Third number , (a + d)  =  25/2 + 5/2 = 25/2 + 5/2 = (25 + 5)/2 = 30/2 = 15

 Fourth number , (a + 3d) =  25/2 + 3(5/2) = 25/2 + 15/2 = (25 +15)/2 = 40/2 = 20/2 = 20

Hence, the four numbers in A.P are 5, 10, 15 and 20.

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Answer 2

Answer: 5, 10, 15, 20

Step-by-step explanation:

Let the four numbers be = a, a+d, a+2d, a+3d

Now, we're given that:

The greatest number is 4 times the least

Greatest number here is = a + 3d

Least number here is = a

We get that:

a + 3d = 4a

Solve this formed equation further:

=> 3a = 3d

Solve it further:

=> a = d

We're provided with the sum:

$\tt{s_{4} = 50}$

We know that:

=> $\tt{s_{n} = \frac{n}{2} (2a + (n-1)d)}$

Put the values and solve this further:

=> $\tt{50 = \frac{4}{2} (2a + 3d)}$

We found that:

a = d

=> $\tt{50 = 2 (2a + 3a)}$

=> 25 = 5a

=> a = 5

a = d = 5

Thus, the AP of four numbers is = 5, 10, 15, 20