Question

Column I	Column II
(a) `4, 1, 1/4, 1/16`	(i) A.P
(b) 2, 3, 5, 7	(ii) Sequence
(c) 13, 8, 3, –2, –7	(iii) G.P.

Answer 1

Column I	Column II
(a) `4, 1, 1/4, 1/16`	(i) G.P
(b) 2, 3, 5, 7	(ii) Sequence
(c) 13, 8, 3, –2, –7	(iii) A.P.

Explanation:

(a) `4, 1, 1/4, 1/16`

Here, `a_2/a_1 = 1/4`

`a_3/a_2 = 1/4`

And `a_4/a_3 = (1/16)/(1/4) = 1/4`

Hence it is G.P.

(b) 2, 3, 5, 7

Here a₂ – a₁ = 3 – 2 = 1

a₃ – a₂ = 5 – 3 = 2

∴ a₂ – a₁ ≠ a₃ – a₂

Hence it is not A.P

`a_2/a_1 = 3/2, a_3/a_2 = 5/3`

So, `3/2 ≠ 5/3`

So it is not G.P.

Hence it is sequence

(c) 13, 8, 3, – 2, – 7

Here a₂ – a₁ = 8 – 13 = – 5

a₃ – a₂ = 3 – 8 = – 5

So, a₂ – a₁ = a₃ – a₂ = – 5

So, it is an A.P.