# 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

c^{2} = bd

c^{2} = ((a + c)/2)d [from (1)]

d = 2c^{2}/(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 / [2c^{2}/(a + c)] = 1/c + 1/e [from (2)]

⇒ 2(a + c)/2c^{2} = 1/c + 1/e

⇒ (a + c)/c^{2} = (e + c)/ec

⇒ (a + c)/c = (e + c)/e

⇒ (a + c)e = (e + c)c

⇒ ae + ce = ec + c^{2}

⇒ c^{2} = 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