If 1/a, 1/b, 1/c are in A.P show that b= 2ac/a+c
Answers
EXPLANATION.
1/a, 1/b, 1/c --------(A.P.).
As we know that,
First term = a = 1/a.
Common difference = d = b - a = c - b.
Common difference = d = 1/b - 1/a.
As we know that,
Conditions of an A.P.
⇒ 2b = a + c.
⇒ 2(1/b) = 1/a + 1/c.
⇒ 2/b = c + a/ac.
⇒ 2ac = b(c + a).
⇒ b = 2ac/a + c.
Hence proved.
MORE INFORMATION.
General term of H.P.
General term of an H.P.
1/a + 1/a + d + 1/a + 2d + ,,,,,,, is
Tₙ = 1/a + (n - 1)d.
Note :
Sum of n terms in H.P. is not defined.
Harmonical Mean (H.M.).
If H is the H.M between a and b then H = 2ab/a + b.
Answer:
Hope it helps you
Step-by-step explanation:
Question:
If 1/a, 1/b, 1/c are in A.P show that b= 2ac/a+c
Answer:
We know that, 1/a, 1/b, 1/c are in A.P
So, the common difference is:
a2-a1 = a3-a2
1/b - 1/a = 1/c - 1/b
1/b+1/b = 1/c + 1/a
2/b = 1/c + 1/a
On taking LCM:
2/b = a+c/ac
b/2 = ac/a+c
b = 2ac/a+c
Hence proved.