JEE MAIN - Mathematics (2019 - 9th January Evening Slot - No. 1)
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 :
2
$${1 \over 2}$$
$${7 \over 13}$$
4
Explanation
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
$${a \over c} = {{A + 6d} \over {A + 12d}} = {{ - 8d} \over { - 2d}} = 4$$
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
$${a \over c} = {{A + 6d} \over {A + 12d}} = {{ - 8d} \over { - 2d}} = 4$$
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