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If a, b, c are in A.P; b, c, d are in G.P and 1/c, 1/d, 1/e, are in A.P. Prove that a, c, e are in G.P
Solution:
It is given that a, b, c are in A.P.
b - a = c - b
b = (a + c)/2 ....(1)
It is given that b, c, d are in G.P
c2 = bd
c2 = ((a + c)/2)d [from (1)]
d = 2c2/(a + c) ....(2)
Also, 1/c, 1/d, 1/e, are in A.P.
Therefore,
⇒ 1/d - 1/c = 1/e - 1/d
⇒ 2/d = 1/c + 1/e
⇒ 2 / [2c2/(a + c)] = 1/c + 1/e [from (2)]
⇒ 2(a + c)/2c2 = 1/c + 1/e
⇒ (a + c)/c2 = (e + c)/ec
⇒ (a + c)/c = (e + c)/e
⇒ (a + c)e = (e + c)c
⇒ ae + ce = ec + c2
⇒ c2 = ae
Thus, a, c, e are in G.P
NCERT Solutions Class 11 Maths Chapter 9 Exercise ME Question 20
If a, b, c are in A.P; b, c, d are in G.P and 1/c, 1/d, 1/e, are in A.P. Prove that a, c, e are in G.P
Summary:
Given that a, b and c are in A.P, b,c and D in G.P and 1/c, 1/d and 1/e in A.P and we proved that a,c and e are in G.P
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