Answer: Option A. -> 44

Answer: Option C. -> 3
$$\eqalign{
& {S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right], \cr
& {S_{2n}} = \frac{{2n}}{2}\left[ {2a + \left( {2n - 1} \right)d} \right]\,{\text{and}} \cr
& {S_{3n}} = \frac{{3n}}{2}\left[ {2a + \left( {3n - 1} \right)d} \right] \cr
& {\text{Now}}\,{S_{2n}} - {S_n} \cr} $$
$$ = \frac{{2n}}{2}\left[ {2a + \left( {2n - 1} \right)d} \right] - $$     $$\frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$$
$$ = \frac{n}{2}\left[ {4a + \left( {4n - 2} \right)d} \right] - $$     $$\left[ {2a + \left( {n - 1} \right)d} \right]$$
$$\eqalign{
& = \frac{n}{2}\left[ {4a - 2a + \left( {4n - 2 - n + 1} \right)d} \right] \cr
& = \frac{n}{2}\left[ {2a + \left( {3n - 1} \right)d} \right] \cr
& = \frac{1}{3}\left( {{S_{3n}}} \right) \cr
& \therefore {S_{3n}}:\left( {{S_{2n}} - {S_n}} \right) \cr
& = 3:1\,{\text{or}}\,\frac{3}{1} = 3 \cr} $$

Answer: Option A. -> -1
$$\eqalign{
& {\text{A}}{\text{.P}}{\text{.}}\,{\text{is}}\,\frac{1}{{2q}},\,\frac{{1 - 2q}}{{2q}},\,\frac{{1 - 4q}}{{2q}},.... \cr
& \Rightarrow \frac{1}{{2q}},\,\left( {\frac{1}{{2q}} - 1} \right),\,\left( {\frac{1}{{2q}} - 2} \right),\,.... \cr
& {\text{Clearly}}\,d = \left( {\frac{1}{{2q}} - 1} \right) - \frac{1}{{2q}} \cr
& = \frac{1}{{2q}} - 1 - \frac{1}{{2q}} \cr
& = - 1 \cr} $$

Answer: Option C. -> 1
$$\eqalign{
& \frac{1}{{x + 2}},\,\frac{1}{{x + 3}},\,\frac{1}{{x + 5}}\,{\text{are}}\,{\text{in}}\,{\text{A}}{\text{.P}}{\text{.}} \cr
& \therefore \frac{1}{{x + 3}} - \frac{1}{{x + 2}} = \,\frac{1}{{x + 5}} - \frac{1}{{x + 3}} \cr
& \Rightarrow \frac{{x + 2 - x - 3}}{{\left( {x + 3} \right)\left( {x + 2} \right)}} = \frac{{x + 3 - x - 5}}{{\left( {x + 5} \right)\left( {x + 3} \right)}} \cr
& \Rightarrow \frac{{ - 1}}{{\left( {x + 3} \right)\left( {x + 2} \right)}} = \frac{{ - 2}}{{\left( {x + 5} \right)\left( {x + 3} \right)}} \cr
& \Rightarrow \frac{{ - 1}}{{x + 2}} = \frac{{ - 2}}{{x + 5}} \cr
& \Rightarrow - 2x - 4 = - x - 5 \cr
& \Rightarrow - 2x + x = - 5 + 4 \cr
& \Rightarrow - x = - 1 \cr
& \therefore x = 1 \cr} $$

Answer: Option B. -> $$\frac{{b - a}}{{n - 1}}$$
In the given A.P.
First term = a and nth term = b
$$\eqalign{
& \therefore a + \left( {n - 1} \right)d = b \cr
& \Rightarrow \left( {n - 1} \right)d = b - a \cr
& \Rightarrow d = \frac{{b - a}}{{n - 1}} \cr} $$

Answer: Option B. -> 7
Sum of 5 + 9 + 13 + . . . . to n terms
$$\eqalign{
& = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] \cr
& {\text{Here}}\,a = 5,\,d = 9 - 5 = 4 \cr
& \therefore {\text{Sum}} = \frac{n}{2}\left[ {2 \times 5 + \left( {n - 1} \right) \times 4} \right] \cr
& = \frac{n}{2}\left[ {10 + 4n - 4} \right] \cr
& = \frac{n}{2}\left[ {6 + 4n} \right] \cr
& = n\left( {3 + 2n} \right) \cr} $$
and sum of 7 + 9 + 11 + . . . . to (n + 1) terms
$$\eqalign{
& = \frac{{n + 1}}{2}\left[ {2 \times 7 + \left( {n + 1 - 1} \right)2} \right] \cr
& = \frac{{n + 1}}{2}\left[ {14 + 2n} \right] \cr
& = \left( {n + 1} \right)\left( {7 + n} \right) \cr
& \therefore \frac{{5 + 9 + 13 + ...\,{\text{to}}\,n\,{\text{terms}}}}{{7 + 9 + 11 + ...\,{\text{to}}\,\left( {n + 1} \right)\,{\text{terms}}}} = \frac{{17}}{{16}} \cr
& \Rightarrow \frac{{n\left( {3 + 2n} \right)}}{{\left( {n + 1} \right)\left( {7 + n} \right)}} = \frac{{17}}{{16}} \cr
& \Rightarrow 16n\left( {3 + 2n} \right) = 17\left( {n + 1} \right)\left( {7 + n} \right) \cr
& \Rightarrow 48n + 32{n^2} = 17\left( {{n^2} + 8n + 7} \right) \cr
& \Rightarrow 48n + 32{n^2} = 17{n^2} + 136n + 119 \cr
& \Rightarrow 48n + 32{n^2} - 17{n^2} - 136n - 119 = 0 \cr
& \Rightarrow 15{n^2} - 88n - 119 = 0 \cr
& \Rightarrow 15{n^2} - 105n + 17n - 119 = 0 \cr} $$
\[\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ \begin{array}{l}
∵ 15 \times \left( { - 119} \right) = 1785\\
- 1785 = 17 \times \left( {105} \right)\\
- 88 = 17 - 105
\end{array} \right\}\]
$$\eqalign{
& \Rightarrow 15n\left( {n - 7} \right) + 17\left( {n - 7} \right) = 0 \cr
& \Rightarrow \left( {n - 7} \right)\left( {15n + 17} \right) = 0 \cr} $$
Either $$n - 7 = 0,$$   then $$n = 7$$
or $$15n + 13 = 0,$$    then $$n = \frac{{ - 13}}{{15}}$$   which is not possible being fraction
∴ n = 7

Answer: Option D. -> None of these
$$\eqalign{
& {a_n} = a + \left( {n - 1} \right)d \cr
& {a_9} = 449 \cr
& \,\,\,\,\,\, = a + \left( {9 - 1} \right)d \cr
& \,\,\,\,\,\, = a + 8d\,.....\left( 1 \right) \cr
& {a_{449}} = 9 \cr
& \,\,\,\,\,\,\,\,\, = a + \left( {449 - 1} \right)d \cr
& \,\,\,\,\,\,\,\,\, = a + 448d\,.....\left( 2 \right) \cr
& {\text{Subtracting}} \cr
& 440d = - 440 \cr
& \Rightarrow d = \frac{{ - 440}}{{440}} = - 1 \cr
& {\text{and}}\,a + 8d = 449 \cr
& \Rightarrow a \times 8 \times \left( { - 1} \right) = 449 \cr
& \Rightarrow a = 449 + 8 = 457 \cr
& \therefore 0 = a + \left( {n - 1} \right)d \cr
& \Rightarrow 0 = 457 + \left( {n - 1} \right)\left( { - 1} \right) \cr
& \Rightarrow 0 = 457 - n + 1 \cr
& \Rightarrow n = 458 \cr
& \therefore {458^{{\text{th}}}}\,{\text{term}} = 0 \cr} $$

Answer: Option B. -> d = 2a
Sn is the sum of first n terms a is the first term and d is the common difference
$$\eqalign{
& {S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] \cr
& \frac{{{S_x}}}{{{S_{kx}}}} = \frac{{\frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]}}{{\frac{{kx}}{2}\left[ {2a + \left( {kx - 1} \right)d} \right]}} \cr
& \because \frac{{{S_x}}}{{{S_{kx}}}}\,{\text{is}}\,{\text{independent of}}\,x \cr} $$
$$\therefore \frac{{\frac{n}{2}\left[ {2a + \left( {x - 1} \right)d} \right]}}{{\frac{{kx}}{2}\left[ {2a + \left( {kx - 1} \right)d} \right]}}\,$$    is independent of x
$$\therefore \frac{{\frac{n}{2}\left[ {2a + xd - d} \right]}}{{\frac{{kx}}{2}\left[ {2a + kdx - d} \right]}}$$    is independent of x
$$ \Rightarrow \frac{{2a - d}}{{k\left( {2a - d} \right)}}$$   is in dependent of x if 2a - d $$ \ne $$ 0
If 2a - d =0, then d = 2a

Answer: Option A. -> $$\frac{{179}}{{321}}$$
Let a1, d2 be the first terms of two ratios S and S' and d1, d2 be their common difference respectively
Then,
$$\eqalign{
& {S_n} = \frac{n}{2}\left[ {2{a_1} + \left( {n - 1} \right){d_1}} \right]\,{\text{and}} \cr
& S{'_n} = \frac{n}{2}\left[ {2{a_2} + \left( {n - 1} \right){d_2}} \right] \cr
& {\text{Now,}} \cr
& \,\frac{{{S_n}}}{{S{'_n}}} = \frac{{\frac{n}{2}\left[ {2{a_1} + \left( {n - 1} \right){d_1}} \right]}}{{\frac{n}{2}\left[ {2{a_2} + \left( {n - 1} \right){d_2}} \right]}} \cr
& = \frac{{2{a_1} + \left( {n - 1} \right){d_1}}}{{2{a_2} + \left( {n - 1} \right){d_2}}} \cr
& {\text{But}}\,\frac{{{S_n}}}{{S{'_n}}} = \frac{{5n + 4}}{{9n + 6}} \cr
& \therefore \frac{{2{a_1} + \left( {n - 1} \right){d_1}}}{{2{a_2} + \left( {n - 1} \right){d_2}}} = \frac{{5n + 4}}{{9n + 6}} \cr} $$
Now we have to find the ratios in 18th term
Here n = 18
$$\eqalign{
& \therefore \frac{{2{a_1} + \left( {18 - 1} \right){d_1}}}{{2{a_2} + \left( {18 - 1} \right){d_2}}} = \frac{{5\left( {2n - 1} \right) + 4}}{{9\left( {2n - 1} \right) + 6}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{5\left( {2 \times 18 - 1} \right) + 4}}{{9\left( {2 \times 18 - 1} \right) + 6}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{5 \times 35 + 4}}{{9 \times 35 + 6}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{175 + 4}}{{315 + 6}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{179}}{{321}} \cr} $$

Answer: Option C. -> 400
First 20 odd natural numbers are 1, 3, 5, 7, 9, 11, 13, 15, . . . . ., 39
Here a = 1, d = 2, n = 20
$$\eqalign{
& \therefore {S_{20}} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] \cr
& = \frac{{20}}{2}\left[ {2 \times 1 + \left( {20 - 1} \right) \times 2} \right] \cr
& = 10\left( {2 + 38} \right) \cr
& = 10 \times 40 \cr
& = 400 \cr} $$