Series Pattern
2 x 3 + 5 = 11, 11 x 4 - 6 = 38
38 x 5 + 7 = 197, 197 x 6 - 8 = 1174
Should come in place of 1172
1174 x 7 + 9 = 8827

Correct Option: D

Series Pattern
2 x 3 + 5 = 11, 11 x 4 - 6 = 38
38 x 5 + 7 = 197, 197 x 6 - 8 = 1174
Should come in place of 1172
1174 x 7 + 9 = 8827
8827 x 8 - 10 = 65806
Clearly, 1172 is wrong and will be replaced by 1174.

Series Pattern
(15 x 11), (15 x 13), (15 x 17), (15 x 19), (15 x 23), (15 x 29)

Correct Option: C

Series Pattern
(15 x 11), (15 x 13), (15 x 17), (15 x 19), (15 x 23), (15 x 29)
∴ Missing term = (15 x 23) = 345

The 1st term of an HP is 1/17 .
Hence, for the corresponding AP, the terms is 17 .
Here, a = 17
nth term of an AP is given by T_n = a + (n -1)d
∴ T₂ = 17 + d, T₄ = 17 + 3d
T₅ = 17 + 4d, T₆ = 17 + 5d
∴ (17 + d) (17 + 3d) = (17 + 4d) (17 + 5d)

Correct Option: A

The 1st term of an HP is 1/17 .
Hence, for the corresponding AP, the terms is 17 .
Here, a = 17
nth term of an AP is given by T_n = a + (n -1)d
∴ T₂ = 17 + d, T₄ = 17 + 3d
T₅ = 17 + 4d, T₆ = 17 + 5d
∴ (17 + d) (17 + 3d) = (17 + 4d) (17 + 5d)
⇒ 289 + 51d + 17d + 3d² = 289 + 85d + 68d + 20d²
⇒ 68d + 3d² = 153d + 20d²
⇒ 17d² = -85d
⇒ d = -5
∴ 3rd term is T₃ = a + 2d = 17 + 2(-5)
= 17 - 10 = 7
Hence, for the corresponding HP, the 3rd term is 1/7.

Let a be the first term and and r be the common ratio of the GP. From the given problem,
a + ar^{n - 1} = 66 ...(i)
Also, ar x ar^{n - 2} = 128 ⇒ a²r^{n - 1} = 128 ... (ii)
From Eq. (ii),
a.ar^{n - 1} = 128
ar^{n - 1} = 128/ a
On substitution this in Eq.(i), we get
a + 128/a = 66

Correct Option: B

Let a be the first term and and r be the common ratio of the GP. From the given problem,
a + ar^{n - 1} = 66 ...(i)
Also, ar x ar^{n - 2} = 128 ⇒ a²r^{n - 1} = 128 ... (ii)
From Eq. (ii),
a.ar^{n - 1} = 128
ar^{n - 1} = 128/ a
On substitution this in Eq.(i), we get
a + 128/a = 66
a² - 66a + 128 = 0
a = [-b ± √b² - 4ac] /2a
= 66 ± √66² - 4 x 128 x 1)/2 = 64 or 2

Put n = 1, now the sum of the series = 1.2.4 = 8
Put n = 1, in the options
(a) 1(1 + 1) (1 + 2) = 6
(b) 1(1 + 1)/12 x (3 + 19 + 26) = 8
(c) (1 + 1) (1 + 2) (1 + 3)/4 = 6
(d) 1(1 + 1) (1 + 2) (1 + 3 )/3 = 8
As, the sum of series = 8
Hence, option (a) and (c) can be rejected.
Now, put n = 2

Correct Option: B

Put n = 1, now the sum of the series = 1.2.4 = 8
Put n = 1, in the options
(a) 1(1 + 1) (1 + 2) = 6
(b) 1(1 + 1)/12 x (3 + 19 + 26) = 8
(c) (1 + 1) (1 + 2) (1 + 3)/4 = 6
(d) 1(1 + 1) (1 + 2) (1 + 3 )/3 = 8
As, the sum of series = 8
Hence, option (a) and (c) can be rejected.
Now, put n = 2
Sum 1. 2. 4 + 2 . 3 . 5 = 38
Put n = 2, in option (b)
2(2 + 1)/12 (3 x 4 + 19 x 2 + 26)38
Hence, (b) is the correct option.