Let term = l = ar^{n - 1} a = 5 and l = 20480
r = 20/5 = 4
∴ 20480 = 5 x (4)^n-1

Correct Option: A

Let term = l = ar^{n - 1} a = 5 and l = 20480
r = 20/5 = 4
∴ 20480 = 5 x (4)^n-1
(4)^n-1 = 20480/5 = 4096 = (4)⁶
n - 1 = 6
∴ n = 7

(a + b)/2 = 25
a + b = 50
√ab = 7
ab = 49

Correct Option: B

(a + b)/2 = 25
a + b = 50
√ab = 7
ab = 49
Hence, A can either be 7 or 49.
So, 49 is the answer.

The second and fourth term of the HP are 1/6 and 1/14 respectively,
Hence, for the corresponding AP, the second term is 6 and fourth term is 14.
Hence a + d = 6 and a + 3d = 14
∴ 2d = 8
⇒ d = 4 and a = 2

Hence, the 10th term of this AP = a + (10 -1)d

Correct Option: C

The second and fourth term of the HP are 1/6 and 1/14 respectively,
Hence, for the corresponding AP, the second term is 6 and fourth term is 14.
Hence a + d = 6 and a + 3d = 14
∴ 2d = 8
⇒ d = 4 and a = 2

Hence, the 10th term of this AP = a + (10 -1)d
= 2 + (10 - 1)x 4 = 2 + 9 x 4
= 2 + 36 = 38
Hence, for the corresponding HP, the 10th term is 1/38.

T₅ = a + 4d, T₇ = a + 6d
∴ 5(a + 4d) = 7(a + 6d)
⇒ 5a + 20d = 7a + 42d
⇒ a = -11d

Correct Option: B

T₅ = a + 4d, T₇ = a + 6d
∴ 5(a + 4d) = 7(a + 6d)
⇒ 5a + 20d = 7a + 42d
⇒ a = -11d
⇒ T₁₂ = a + 11d = - 11d + 11d = 0
So, the twelfth term is 0.

Series pattern
2 x 1², 2 x 2², 2 x 3², 2 x 4², 2 x 5², 2 x 6²

Correct Option: D

Series pattern
2 x 1², 2 x 2², 2 x 3², 2 x 4², 2 x 5², 2 x 6²
∴ Missing term = 2 x 4² = 32