Solution:
211X
Go through option C
(46)² = 2116

Solution:
$$\eqalign{ 23)\,\,\,1056\,\,\,\,(45 \cr \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,92 \cr \,\,\,\,\,\,\,\,\,\,\, - - - \cr \,\,\,\,\,\,\,\,\,\,\,\,\,\,136 \cr \,\,\,\,\,\,\,\,\,\,\,\,\,\,115 \cr \,\,\,\,\,\,\,\,\,\,\, - - - \cr \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,21 \cr \,\,\,\,\,\,\,\,\,\,\, - - - \cr {\text{Required}}\,{\text{number}} \cr = \left( {23 - 21} \right) \cr = 2 \cr} $$

Solution:
The required arithmetic sequence of five prime numbers is 5, 11, 17, 23, 29 and therefore, the required 5^th term is 29

Solution:
$$\eqalign{ {\text{Let number are }}x,{\text{ }}x + 5,{\text{ }}x + 10 \cr {\text{So, }}x{\text{ + }}x + 5 + x + 10 = 225 \cr 3x + 15 = 225 \cr 3x = 210 \cr x = 70 \cr {\text{largest number is }}70 + 10 = 80 \cr} $$

Solution:
$$\eqalign{ {\text{The}}\,{\text{given}}\,{\text{expression}}\,{\text{is}}\,{\text{in}}\,{\text{GP}}\,{\text{series}} \cr S = {3^0} + {3^1} + {3^2} + {3^3} + ........ + {3^{200}} \cr S = {\frac{{ {{3^0} \times \left( {{3^{201}} - 1} \right)} }}{{ {3 - 1} }}} \cr S = \frac{{ {{3^{201}} - 1} }}{2} \cr S = \frac{{ {{{\left( {{3^3}} \right)}^{67}} - {1^3}} }}{2} \cr S = \frac{{ {{{27}^{67}} - {1^3}} }}{2} \cr} $$
Since, (Aⁿ - Bⁿ) is divisible by (A - B), So, (27⁶⁷ - 1³) is divisible by (27 - 1) = 26
Hence, Expression is also divisible by 13 as it is divisible by 26
Thus given expression is divisible by 13 so the remainder will be 0

Solution:
37 × 3 is divisible by 7
⇒ (7 × 3 - 3) is either 0 or divisible by 7
⇒ 7 × 0 is divisible by 7
⇒ X = 0 or X = 7

Solution:
2^m > 960 when the least value of m is 10
Then, 2¹⁰ = 1024 and 1024 - 960 = 64 = 2⁶
∴ m = 10 and n = 6