let p,q,r be positive integers such that q/p is an integer if p,q,r are in geometric progression and the arithmetic mean of p,q,r is q+5 also
x = p2-p+14/p+1
then[x] is where [.] denotes greater integer function
Answers
EXPLANATION.
Let, p, q, r be positive integers.
Such that q/p is an integers.
p, q, r are in geometric progression.
The arithmetic mean of p, q, r is q + 5.
As we know that,
⇒ p, q, r - - - - - (G.P).
⇒ p = a.
⇒ q = ar.
⇒ r = ar².
⇒ (p + q + r)/3 = (q + 5).
⇒ (p + q + r) = 3(q + 5).
⇒ (p + q + r) = 3(q) + 15.
⇒ a + ar + ar² = 3(ar) + 15.
⇒ a + ar + ar² - 3ar = 15.
⇒ a - 2ar + ar² = 15.
⇒ a(1 - 2r + r²) = 15.
⇒ (1 - 2r + r²) = 15/a.
⇒ (r² - 2r + 1) = 15/a.
⇒ (r - 1)² = 15/a.
Since, 15/a must be perfect square and a ∈ N.
So, a = 15.
Put the value of a = 15 in the equation, we get.
⇒ (r - 1)² = 15/15.
⇒ (r - 1)² = 1.
⇒ (r - 1) = ± 1.
⇒ r = 0, 2.
⇒ r = 2.
⇒ a = p = 15.
To find :
⇒ x = (p² - p + 14)/p + 1.
⇒ x = [(15)² - (15) + 14]/(15 + 1).
⇒ x = [225 - 15 + 14]/(16).
⇒ x = [239 - 15]/(16).
⇒ x = [224]/(16).
⇒ x = 14.
Option [B] is correct answer.
Given :-
Let p,q,r be positive integers such that q/p is an integer if p,q,r are in geometric progression
To Find :-
x = p² - p + 14/p + 1
Solution :-
We know that
Terms of GP = a, ar, ar², ar³
Now
p = a
q = ar
r = ar²
Arthimetic Mean = a + 5
Now
p + q + r/3 = a + 5
p + q + r = 3(a + 5)
p + q + r = 3a + 15
By putting the value
a + ar + ar² = 3ar + 15
a + ar + ar² - 3ar = 15
a + (ar - 3ar) + ar² = 15
a + (-2ar) + ar² = 15
a - 2ar + ar² = 15
Taking a as common
a[1 - 2(1)r + 1(r²)] = 15
a[1 - 2r + r²] = 15
1 - 2r + r² = 15/a
(r - 1)² = 15/a
Value of p = 15
(r - 1)² = 15/15
(r - 1)² = 1
√(r - 1)² = √1
r - 1 = ±1
r = 1 ± 1
Either
r = 1 + 1
r = 2
or,
r = 1 - 1
r = 0
Since, they are positive integer. So, we will neglect 0
Finding value of x
x = (15)² - 15 + 14/15 + 1
x = 225 - 15 + 14/16
x = (225 + 14) - 15/16
x = 239 - 15/16
x = 224/16
x = 28/2
x = 14