Question

four numbers are in AP if sum of numbers is 50 and largest number is 4 times the smaller one then find the numbers​

Answer 1

Given :

Four numbers are in an A.P

Let the numbers be a + a + d + a + 2d + a + 3d

Sum of those four numbers = 50

According to the problem,

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

4a + 6d = 50 --------(1)

Largest term is 4 times the smaller one.

a + 3d = 4(a)

a + 3d = 4a

3d = 4a - a

3d = 3a

Cancel the number 3.

Then,

d = a

Substitute the value of d in eq - (1)

4a + 6d = 50

4a + 6a = 50

10a = 50

a = 50/10

a = 5

First term = 5

Then,

Second term = a + d = 5 + 5 = 10

Third term = a + 2d = 5 + 10 = 15

Fourth term = a + 3d = 5 + 15 = 20

Therefore, the four terms are 5,10,15 and 20.

Answer 2

Given:

The sum of the four numbers is 50

and fourth term is equal to four times the first term.

General formula of a term of A.P is:-

a+(n-1)d

where a is the first term,d is the common difference and n is the no.of terms

Now,

Fourth term=4 times first term

→a+(4-1)d=4[a+(1-1)d]

→a+3d=4a

→3a=3d

→a=d

The first term and common difference of the A.P. are same

Here,

Sum of the four terms:50

→n/2[2a+(n-1)d]=50

→4/2[2a+(4-1)a]=50

→2(2a+3a)=50

→5a=25

→a=5

→a=d=5

The A.P. would be,

a,a+d,a+2d,.............

5,10,15,.................