Answer: Option D. -> $$\sqrt {112} ,$$
$$\eqalign{
& {\text{A}}{\text{.P}}{\text{.}}\,{\text{is}}\,\sqrt 7 ,\,\sqrt {28} ,\,\sqrt {63} ,\,...... \cr
& \Rightarrow \sqrt 7 ,\,\sqrt {4 \times 7} ,\,\sqrt {9 \times 7} ,\,..... \cr
& \Rightarrow \sqrt 7 ,\,2\sqrt 7 ,\,3\sqrt 7 ,...... \cr
& \therefore Here\,\,a = \sqrt 7 \,{\text{and}} \cr
& d = 2\sqrt 7 - \sqrt 7 = \sqrt 7 \cr
& \therefore {\text{Next}}\,{\text{term}} = 4\sqrt 7 \cr
& = \sqrt {\left( {16 \times 7} \right)} \cr
& = \sqrt {112} \cr} $$

Answer: Option A. -> 44
The third term t3 = a + 2d
The ninth term t9 = a + 8d
t3 + t9 = 2a + 10d = 8
Sum of first 11 terms of an AP is given by
$$\eqalign{
& \Rightarrow {S_{11}} = \frac{{11}}{2}\left[ {2a + 10d} \right] \cr
& \Rightarrow {S_{11}} = \frac{{11}}{2} \times 8 \cr
& \Rightarrow {S_{11}} = 44 \cr} $$

Answer: Option C. -> 5
2 (3y + 5) = 3y – 1 + 5y + 1
(If a, b, c are in A.P., b – a = c – b ⇒ 2b = a + c)
⇒ 6y + 10 = 8y
⇒ 10 = 2y
⇒ y = 5

Answer: Option C. -> 4n + 3
Let a be the first term and d be the common difference of an A.P. and
$$\eqalign{
& {S_n} = 2{n^2} + 5n \cr
& \therefore {S_1} = 2{\left( 1 \right)^2} + 5 \times 1 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\, = 2 + 5 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\, = 7 \cr
& \therefore {S_2} = 2{\left( 2 \right)^2} + 5 \times 2 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 8 + 10 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 18 \cr
& \therefore {\text{First}}\,{\text{term}}\,\left( {{a_1}} \right) = 7\,{\text{and}} \cr
& {\text{Second}}\,{\text{term}}\,{a_2} = {S_2} - {S_1} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 18 - 7 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 11 \cr
& \therefore d = {a_2} - {d_1} \cr
& \,\,\,\,\,\,\,\,\,\,\, = 11 - 7 \cr
& \,\,\,\,\,\,\,\,\,\,\, = 4 \cr
& Now\,{a_n} = a + \left( {n - 1} \right)d \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 7 + \left( {n - 1} \right)4 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 7 + 4n - 4 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4n + 3 \cr} $$

Answer: Option C. -> 89
7th term (a7) = a + 6d = 34
13th term (a13) = a + 12d = 64
Subtracting, 6d = 30 ⇒ d = 5
and a + 12 x 5 = 64
⇒ a + 60 = 64
⇒ a = 64 – 60 = 4
18th term (a18)
= a + 17d
= 4 + 17 x 5
= 4 + 85
= 89

Answer: Option D. -> $$\frac{{n + 1}}{n}$$
Sum of n even natural number = n(n+1)
and sum of n odd natural numbers = n2
$$\eqalign{
& \therefore n\left( {n + 1} \right) = k{n^2} \cr
& \Rightarrow k = \frac{{n(n + 1)}}{{{n^2}}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{n + 1}}{n} \cr} $$

Answer: Option C. -> p + q
In an A.P. Sp = q, Sq = p
Sp+q = Sum of (p + q) terms
= Sum of p term + Sum of q terms
= p + q

Answer: Option C. -> 30
Number of 2-digit positive integers divisible by 4
The smallest 2-digit positive integer divisible by 4 is 12. The largest 2-digit positive integer divisible by 4 is 96.
All the 2-digit positive integers are terms of an Arithmetic progression with 12 being the first term and 96 being the last term.
The common difference is 4.
The nth term an = a1 + (n - 1)d, where a1 is the first term, 'n' number of terms and 'd' the common difference.
So, 96 = 12 + (n - 1) × 4
84 = (n - 1) × 4
Or (n - 1) = 21
Hence, n = 22
i.e., there are 22 2-digit positive integers that are divisible by 4. Number of 2-digit positive integers divisible by 9
The smallest 2-digit positive integer divisible by 9 is 18. The largest 2-digit positive integer divisible by 9 is 99.
All the 2-digit positive integers are terms of an Arithmetic progression with 18 being the first term and 99 being the last term.
The common difference is 9
The nth term an = a1 + (n - 1)d, where a1 is the first term, 'n' number of terms and 'd' the common difference.
So, 99 = 18 + (n - 1) × 9
Or 81 = (n - 1) × 9
Or (n - 1) = 9
Hence, n = 10
i.e., there are 10 2-digit positive integers that are divisible by 9.
Removing double count of numbers divisible by 4 and 9
Numbers such as 36 and 72 are multiples of both 4 and 9 and have therefore been counted in both the groups.
There are 2 such numbers.
Hence, number of 2-digit positive integers divisible by 4 or 9
= Number of 2-digit positive integers divisible by 4 + Number of 2-digit positive integers divisible by 4 - Number of 2-digit positive integers divisible by 4 and 9
= 22 + 10 - 2
= 30

Answer: Option D. -> Both (A) and (B)
Let the numbers are be a - d, a, a + d
Then a - d + a + a + d = 21
3a = 21
a = 7
and (a - d)(a + d) = 45
a2 - d2 = 45
d2 = 4
d = $$ \pm $$ 2
Hence, the numbers are 5, 7 and 9 when d = 2 and 9, 7 and 5 when d = -2. In both the cases numbers are the same.

Answer: Option B. -> 1,64,850
The smallest 3 digit number that will leave a remainder of 2 when divided by 3 is 101.
The next number that will leave a remainder of 2 when divided by 3 is 104, 107, ....
The largest 3 digit number that will leave a remainder of 2 when divided by 3 is 998.
So, it is an AP with the first term being 101 and the last term being 998 and common difference being 3.
Sum of an AP =
$$\left( {\frac{{{\text{First Term}} + {\text{Last Term}}}}{2}} \right) \times $$     $${\text{Number}}\,{\text{of}}\,{\text{Terms}}$$
We know that in an A.P., the nth term an = a1 + (n - 1) × d
In this case, therefore, 998 = 101 + (n - 1) × 3
i.e., 897 = (n - 1) × 3
Therefore, n - 1 = 299
Or n = 300
Sum of the AP will therefore, be
$$\eqalign{
& = \frac{{101 + 998}}{2} \times 300 \cr
& = 1,64,850 \cr} $$