If a2a3/a1a4=a2+a3/a1+a4=3(a2-a4/a1-a4) then a1,a2,a3,a4 are in
A) AP
B) GP
C) HP
D) None of these
Please Answer with Full Explanation
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Answers
Answer:
a)AP is the answer
Step-by-step explanation:
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Answer:
hp
Step-by-step explanation:
we have a2a3/a1a4 = a2+a3/a1+a4
a2+a3/a2a3 = a1+a4/a1a4
therefore,
1/a3 +1/a2 = 1/a4 + 1/a1
1/a2 - 1/a1 = 1/a4 - 1/a3 -------------------- 1
Again,
a2a3/a1a4 = 3(a2-a3/a1-a4)
a1-a4/a1a4 = 3(a2 -a3/a2a3)
(1/a4 - 1/a1) = 3(1/a3 - 1/a2)
(1/a4 - 1/a3) + (1/a3 - 1/a2) + (1/a2 - 1/a1) = 3(1/a3-1/a2
from equation 1
(1/a4 - 1/a3 = 1/a2 - 1/a1)
(1/a2 - 1/a1) + (1/a2 - 1/a1) = 2(1/a3 - 1/a2)
2(1/a2 - 1/a1) = 2(1/a3 - 1/a2)
1/a2 - 1/a1 = 1/a3 - 1/a2 ---------------- 2
from equation 1 and 2
(1/a2 - 1/a1) = (1/a3 - 1/a2) = (1/a4 - 1/a3)
as common difference is same
therefore,
1/a1 , 1/a2 , 1/a3 , 1/a4 are in ap
therefore,
a1 , a2 , a3 ,a4 are in hp