Progressions - Aptitude MCQ Questions and Solutions with Explanations

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

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

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
$$\eqalign{ & \therefore \frac{{2{a_1} + \left( {18 - 1} \right){d_1}}}{{2{a_2} + \left( {18 - 1} \right){d_2}}} = \frac{{5\left( {2n - 1} \right) + 4}}{{9\left( {2n - 1} \right) + 6}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{5\left( {2 \times 18 - 1} \right) + 4}}{{9\left( {2 \times 18 - 1} \right) + 6}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{5 \times 35 + 4}}{{9 \times 35 + 6}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{175 + 4}}{{315 + 6}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{179}}{{321}} \cr} $$

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

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

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 n^th term a_n = 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

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 a_n = 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
$$\eqalign{ & = \frac{{101 + 998}}{2} \times 300 \cr & = 1,64,850 \cr} $$

B is in G.P. with a = 2⁰, r = 2, n = 65
$$\eqalign{ & \therefore {S_n} = \frac{{a\left( {{r^n} - 1} \right)}}{{r - 1}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{2^0}\left( {{2^{65}} - 1} \right)}}{{2 - 1}} \cr & \therefore B = {2^{65}} - 1 \cr & \Rightarrow B = A - 1 \cr & \therefore A\,{\text{is}}\,{\text{larger}}\,{\text{then}}\,B\,{\text{by}}\,1 \cr} $$

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.

Each time the ball is dropped and it bounces back, it reaches $$\frac{{2}}{{3}}$$ of the height it was dropped from.
After the first bounce, the ball will reach $$\frac{{2}}{{3}}$$ of the height from which it was dropped - let us call it the original height. After the second bounce, the ball will reach $$\frac{{2}}{{3}}$$ of the height it would have reached after the first bounce.
So, at the end of the second bounce, the ball would have reached $$\frac{{2}}{{3}}$$ × $$\frac{{2}}{{3}}$$ of the original height = $$\frac{{4}}{{9}}$$ th of the original height.
After the third bounce, the ball will reach $$\frac{{2}}{{3}}$$ of the height it would have reached after the second bounce.
So, at the end of the third bounce, the ball would have reached $$\frac{{2}}{{3}}$$ × $$\frac{{2}}{{3}}$$ × $$\frac{{2}}{{3}}$$ = $$\frac{{8}}{{27}}$$ th of the original height.
After the fourth and last bounce, the ball will reach $$\frac{{2}}{{3}}$$ of the height it would have reached after the third bounce.
So, at the end of the last bounce, the ball would have reached $$\frac{{2}}{{3}}$$ × $$\frac{{2}}{{3}}$$ × $$\frac{{2}}{{3}}$$ × $$\frac{{2}}{{3}}$$ of the original height = $$\frac{{16}}{{81}}$$ of the original height.