There are 4 natural numbers in an increasing order, such that the first three are in an AP and the last three are in a GP. Also the difference between the last and 1st number is 40.
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Given : 4 natural numbers in an increasing order, such that the first three are in an AP and the last three are in a GP. Also the difference between the last and 1st number is 40.
To Find : Numbers
Solution:
Let say 4 numbers are
a , a + d , a + 2d , a + 40
as first three are in an AP and difference between the last and 1st number is 40.
also numbers in an increasing order
=> a + 2d < a + 40
=> 2d < 40
=> d < 20
last three are in a GP
=> (a + 2d)² = ( a + d)( a + 40)
=> a² + 4d² + 4ad = a² + ad + 40a + 40d
⇒ 4d² + 3ad = 40a + 40d
=> 4d(d – 10) = a(40 – 3d)
lets take a = 4d
=> d – 10 = 40 – 3d => d = 50/4
lets take a = 2d
2d – 20 = 40 – 3d => d = 12
a = 24
24 36 48 64
lets take a = d
4d – 40 = 40 – 3d => d = 80
but d < 20
At least we get one set of 4 natural numbers in an increasing order, such that the first three are in an AP and the last three are in a GP. Also the difference between the last and 1st number is 40.
24 36 48 64
24 , 36 and 48 are in AP with d = 12
36 , 48 and 64 are in GP with r = 4/3
64 – 24 = 40
Satisfy all conditions
24 36 48 64
Learn More:
Find the sum of frist 51 terms of the AP whose 2nd term is 2 and 4th …
brainly.in/question/7655866
the 9th and 19th terms of a series of a.p are respectively 35 and 75 …
Given : 4 natural numbers in an increasing order, such that the first three are in an AP and the last three are in a GP. Also the difference between the last and 1st number is 40.
To Find : Numbers
Solution:
Let say 4 numbers are
a , a + d , a + 2d , a + 40
as first three are in an AP and difference between the last and 1st number is 40.
also numbers in an increasing order
=> a + 2d < a + 40
=> 2d < 40
=> d < 20
last three are in a GP
=> (a + 2d)² = ( a + d)( a + 40)
=> a² + 4d² + 4ad = a² + ad + 40a + 40d
⇒ 4d² + 3ad = 40a + 40d
=> 4d(d – 10) = a(40 – 3d)
lets take a = 4d
=> d – 10 = 40 – 3d => d = 50/4
lets take a = 2d
2d – 20 = 40 – 3d => d = 12
a = 24
24 36 48 64
lets take a = d
4d – 40 = 40 – 3d => d = 80
but d < 20
At least we get one set of 4 natural numbers in an increasing order, such that the first three are in an AP and the last three are in a GP. Also the difference between the last and 1st number is 40.
24 36 48 64
24 , 36 and 48 are in AP with d = 12
36 , 48 and 64 are in GP with r = 4/3
64 – 24 = 40
Satisfy all conditions
24 36 48 64
Learn More:
Find the sum of frist 51 terms of the AP whose 2nd term is 2 and 4th …
brainly.in/question/7655866
the 9th and 19th terms of a series of a.p are respectively 35 and 75 …
brainly.in/question/7568849