Progressions - Aptitude MCQ Questions and Solutions with Explanations

Answer: Option D

$$\eqalign{ & {a_n} = a + \left( {n - 1} \right)d \cr & {a_9} = 449 \cr & \,\,\,\,\,\, = a + \left( {9 - 1} \right)d \cr & \,\,\,\,\,\, = a + 8d\,.....\left( 1 \right) \cr & {a_{449}} = 9 \cr & \,\,\,\,\,\,\,\,\, = a + \left( {449 - 1} \right)d \cr & \,\,\,\,\,\,\,\,\, = a + 448d\,.....\left( 2 \right) \cr & {\text{Subtracting}} \cr & 440d = - 440 \cr & \Rightarrow d = \frac{{ - 440}}{{440}} = - 1 \cr & {\text{and}}\,a + 8d = 449 \cr & \Rightarrow a \times 8 \times \left( { - 1} \right) = 449 \cr & \Rightarrow a = 449 + 8 = 457 \cr & \therefore 0 = a + \left( {n - 1} \right)d \cr & \Rightarrow 0 = 457 + \left( {n - 1} \right)\left( { - 1} \right) \cr & \Rightarrow 0 = 457 - n + 1 \cr & \Rightarrow n = 458 \cr & \therefore {458^{{\text{th}}}}\,{\text{term}} = 0 \cr} $$

Answer: Option B

S_n 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

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 18^th 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

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

Answer: Option D

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