Let a, b and c be the 7th, 11th and 13th terms respectively of a non-constant A.P.

Question

Let a, b and c be the 7th, 11th and 13th terms respectively of a non-constant A.P.  If these are also three consecutive terms of a G.P., then $${a \over c}$$ equal to :

Accepted Answer

T7 = A + 6d = a; T11 = A + 10d = b; T13 = A + 12d = c

Now a, b, c are in G.P.

$$ \therefore $$  b2 = ac

$$ \Rightarrow $$  (A + 10d)2 = (A + 6d) (A + 12d)

$$ \Rightarrow $$  A2 + 100d2 + 20Ad = A2 + 18Ad + 72d2

$$ \Rightarrow $$  A + 14d = 0, A = - 14d

$${a \over c} = {{A + 6d} \over {A + 12d}} = {{ - 8d} \over { - 2d}} = 4$$

Solutions · 1