find 4 integers a,b,c,d such that a,b,c are in GP; b,c,d are in AP and c+d=20,a+b=6
Answers
Given : 4 integers a,b,c,d such that a,b,c are in GP b,c,d are in AP and c+d=20,a+b=6
To find : 4 integers a,b,c,d
Solution:
Let say a , b , c are in GP
=> b² = ac
b , c , d are in AP
=> b + d = 2c
a + b = 6
c + d = 20
a + b + c + d = 26
=> a + c + b + d = 26
=> a + c + 2c = 26
=> a + 3c = 26
=> c = (26 - a)/3
b² = ac
=> b² = a(26 - a)/3
=> 3b² = 26a - a²
a + b = 6
=> b = 6 - a
=> 3(6 - a)² = 26a - a²
=> 3(a² - 12a + 36) = 26a - a²
=> 3a² - 36a + 108 = 26a - a²
=> 4a² - 62a + 108 = 0
=> 2a² - 31a + 54 = 0
=> 2a² - 4a - 27a + 54 = 0
=> 2a(a - 2) - 27(a - 2) = 0
=> (2a - 27)(a - 2) = 0
=> a = 2 , a = 27/2
27/2 is not an integer
a = 2
=> b = 6 - 2 = 4
b² = ac
=> 4² = 2(c)
=> c = 8
c + d = 20
=> 8 + d = 20
=> d = 12
2 , 4 , 8 & 12 are the integers
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Answer:
how to b²=ac
Step-by-step explanation:
and how to
b+d=2c