If a, b, c are in A.P. then prove that
are also in A.P.
Given, a, b, c are in A.P.
Here, first term = a
Common difference, d1 = b – a … (1)
and d2 = c – b … (2)
Consider
,
and
,
Common difference, d3 =
- ![]()
= ![]()
=
… (3)
⇒ d4 =
- ![]()
= ![]()
=
… (4)
From (1) and (2),
⇒ d1 = d2
⇒ b – a = c – b
Dividing both sides by abc,
⇒
= ![]()
⇒ d3 = d4 [From (3) and (4)]
Hence,
,
and
are in A.P.
Couldn't generate an explanation.
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