XAT Progressions and Series Questions

Given that $$a_1=1$$ & $$a_{n+1} =\frac{1}{1+\frac{1}{a_{n}}}$$

$$a_2=\ \frac{\ 1}{1+\ \frac{\ 1}{1}}=\ \ \frac{1}{2},\ a_3=\ \frac{\ 1}{1+\ \frac{\ 1}{\left(\frac{1}{2}\right)}}=\frac{1}{3},....$$

This implies, $$\ a_n=\ \frac{1}{n}$$.

Required value$$=a_1a_2+a_2a_3+........+a_{2008}a_{2009}$$

= $$\frac{1}{1}\times\ \frac{1}{2}+\frac{1}{2}\times\ \frac{1}{3}+.....+\frac{1}{2008}\times\ \frac{1}{2009}$$

= $$\left(\frac{1}{1}-\ \frac{1}{2}\right)+\left(\frac{1}{2}-\ \frac{1}{3}\right)+.....+\left(\frac{1}{2008}-\ \frac{1}{2009}\right)$$

= $$1-\ \frac{1}{2009}$$ 

= $$\frac{2008}{2009}$$

The answer is option C.

Given the ball falls from a height of 3 meters.

The ball reaches a height which 0.8 times the original height every time.

Hence this is in the form of a geometric progression. We need to count distance when the ball flies upward and downward.

Hence considering every time the ball flies upward to a series with terms :

h1, h2,..........................

Every time the ball falls down to be

d1, d2 ,...............

h1 = (0.8)*3, h2 = (0.8)*(0.8)*3 ,.........................

d1 = 3, d2 =3*(0.8), d3 = 3*(0.8)*(0.8)................

h1+h2 ........ = Sum of an infinite geometric progression. = 3*0.8(1+0.8+$$0.64+...............$$)

THe sum of an infinite GP with r less than 1 is : $$\frac{a}{1-r}$$

= $$2.4\cdot\left(\frac{1}{1-0.8}\right)\ =\ 12\ meters$$

The sum of d1+d2++........................

= 3 + (h1+h2+..................) = 15.

The total distance = 15 + 12 = 27 meters.

We have 
3.1416141614161416......
=3+0.1416+0.00001416+0.000000001416.....
= 3+$$\frac{1416}{10^4}+\frac{1416}{10^8}+\frac{1416}{10^{12}}+......$$
Now excluding 3 we get a series with infinite Geometric progression such that first term is $$\frac{1416}{10^4}$$ and common ratio is $$\frac{1}{10^4}$$
Therefore we get sum as $$\frac{a}{1-r}$$
=$$\frac{1416}{\frac{10^4}{1-\frac{1}{10^4}}}$$
we get sum as $$\frac{1416}{10^4-1}$$
=$$\frac{1416}{9999}$$
Now adding 3 we get value as :
$$\frac{31413}{9999}$$
Now taking 3 common we  get ratio of A:B as
$$\frac{10471}{3333}$$
So A-B will be = 7138

The two given sequences in AP are : 

x, a1, a2, y and x, b1, b2, z.

Additionally, it is given that : y  > x and z < x.

Hence the common difference is not zero for both the series :

Since y > x the common difference is positive for the first series. (Considering the common difference to be d1)

Similarly z < x the common difference is negative for the given series. (Considering the common difference to be d2)

Now for the given value :

$$\frac{\left(a1-a2\right)}{\left(b1-b2\right)}$$

The value of a1 - a2 is negative and b1 - b2 is positive.

Hence the value of $$\frac{\left(a1-a2\right)}{\left(b1-b2\right)}$$ takes a negative value.

The only possible option is -3.

The answer is option C.

The walkway is 200 meters long, and stones are placed 5 meters apart on both sides, starting at 0 meters (the start) and ending at 200 meters (the end). The number of stones per side is calculated as follows: from 0 to 200 meters inclusive, with a stone every 5 meters, there are 200/5+1=41 stones per side (0, 5, 10, ..., 195, 200). Since there are two sides, the total number of stones is 41×2=82. However, the first stone on each side (at 0 meters) is laid without walking, as all stones start at the beginning, so we subtract these two initial placements, leaving 82−2=80 stones requiring trips.

The man lays stones in an identical pattern along one side (Side A) and then along the other (Side B). He carries one stone at a time, walks to the spot, lays it, and then walks back to the start to get the next stone. After laying all the stones, he walks back to the start.

Excluding the first stone at 0 meters (no walking), he lays 40 stones (5, 10, ..., 200 meters).
For each stone: Walk to the spot (distance d).
Walk back to the start (distance d)
Total per trip = 2d.

Distances: 5, 10, 15, ..., 200 meters.
This is an arithmetic series: first term = 5, last term = 200, number of terms = 40
The sum of distances to lay stones = 5+10+...+200
The number of terms being 40, the sum will be 20(205)=4100
Total for Side A (to and back): This multiplied by 2 will be 8200

He repeats the exact process for the other side: 40 stones at 5, 10, ..., 200 meters.
Total distance for Side B = 8200 meters (same calculation).

Thus, the total distance the man walks is 16,400 meters. 

The stack of apples can be imagined as shown below. For every row, there is an increase of 1 apple from the previous row.

This means that the sum total of the apples present in the stack will be sum of AP given as : 1,2,3,4.... ie the sum of numbers from 1.

This is given by the formula S = $$\frac{n \times (n+1)}{2}$$

Since the initial number of apples was not more than 150, we find the maximum number of rows that was possible. 

Let S be the total initial sum of apples in the cart. 

We know that $$S\leq 150 $$

$$ \Rightarrow \frac{n \times (n+1)}{2} \leq 150 $$

$$ \Rightarrow n \times (n+1) \leq 300 $$

$$ \Rightarrow n^2 + n \leq 300 $$

Since the biggest number with square less than 300 is 17, we try to see if 17 works in the expression by hit and trial method. 

So, 17 x 18 = 306 > 300, therefore the maximum number of rows possible is not 17.

Since 306 is JUST over 300, let us check at 16.

So, 16 x 17 = 272 < 300, therefore the maximum number of rows possible is 16.

Let us tabulate the total number of apples present in the cart based on the number of rows : 

Since the total number of apples from row 15 is less than 126 (which was found by the customer), we can safely say that the total number of rows of apples with the shopkeeper that day was 16. 

Now, if 'n' rows of apples are sold, the total available apples can be calculated as : $$\frac{16 \times 17}{2} -  \frac{n \times (n+1)}{2}$$

Let us tabulate the result obtained in this case : 

From the table we can see that when the customer saw 126 apples, the apples in the top 4 rows had been sold. 

Therefore, the number of rows = 16 - 4 = 12 rows. 

The given series is (-100)+ (-95)+ (-90)+....+110 + 115 + 120. 
We can observe that -100 will cancel out 100, -95 will cancel out 95 and so on. Therefore, the only terms that will be remaining are 105, 110, 115 and 120.
Sum of the series = 105 + 110 + 115 + 120 = 450.
Therefore, option D is the right answer.

Set :  {$$1, 3, 3^{2}, 3^{3},…...,3^{100}$$}

Clearly, this set is a G.P. with common ratio, $$r = 3$$

Sum of G.P. = $$\frac{a (r^n - 1)}{r - 1}$$

Number of terms = 101

Last term = $$3^{100}$$

Sum of remaining terms = $$\frac{1 (3^{100} - 1)}{3 - 1}$$

= $$\frac{3^{100} - 1}{2}$$

$$\therefore$$ Required ratio = $$\frac{3^{100}}{\frac{3^{100} - 1}{2}}$$

= $$\frac{3^{100} \times 2}{3^{100} - 1}$$

$$\approx \frac{3^{100} \times 2}{3^{100}} = 2$$

The series is an A.P. with common difference, $$d = -66 - (-64) = -2$$

First term, $$a = -64$$ and last term $$a_n = -100$$

nth term of the series, $$a_n = a + (n - 1) d$$

=> $$-100 = -64 + (n - 1) (-2)$$

=> $$n - 1 = \frac{-36}{-2} = 18$$

=> $$n = 18 + 1 = 19$$

$$\therefore$$ Sum = $$\frac{n}{2} (a + a_n)$$

= $$\frac{19}{2} \times (-64 -100) = \frac{19}{2} \times (-164)$$

= $$19 \times (-82) = -1558$$

i. $$a_{741}$$ has 741 ones, it will be divisible by 3

ii. $$a_{534}$$ has 534 ones, it will be divisible by 3

iii. $$a_{123}$$ has 123 ones, it will be divisible by 3

iv. $$a_{77}$$ has 77 ones and each pair of 11 ones is divisible by 13.

Therefore, i,ii,iii and iv are not prime.

Ans -(D) All are correct.

Integers = $$4,4,4,8,10,20,x$$

Clearly, irrespective of the value of $$x$$, Mode = $$4$$

Sum of above integers = $$4 + 4 + 4 + 8 + 10 + 20 + x$$

= $$50 + x$$

Mean = $$\frac{50 + x}{7}$$

Case 1 : If $$x < 4$$

Median of $$x,4,4,4,8,10,20$$ = 4

Mode = 4

Since the median and mode of the dataset are equal, they cannot be arranged in increasing order.

So this case is rejected.

Case 2 : If $$4 < x < 8$$

Median of $$4,4,4,x,8,10,20$$ = $$x$$

Mode = 4

Mean = $$\frac{50 + x}{7}$$

=> $$\frac{54}{7} < Mean < \frac{58}{7}$$

As these are in AP => $$x = 6$$ and Mean = $$8$$

Case 3 : If $$x > 8$$

Mean = $$\frac{50 + x}{7} > \frac{58}{7}$$

Median of $$4,4,4,8,x,10,20$$ = $$8$$

Mode = $$4$$

As these are in AP, => Mean = $$12$$

=> $$\frac{50 + x}{7} = 12$$

=> $$50 + x = 84$$

=> $$x = 84 - 50 = 34$$

$$\therefore$$ Sum of all possible values of x is = $$6 + 34 = 40$$

Length of side of $$S_{n + 1}$$ = Length of diagonal of $$S_n$$

=> Length of side of $$S_{n + 1}$$ = $$\sqrt{2}$$ (Length of side of $$S_{n}$$)

=> $$\frac{\textrm{Length of side of }S_{n + 1}}{\textrm{Length of side of }S_n} = \sqrt{2}$$

=> Sides of $$S_1 , S_2 , S_3 , S_4,........, S_n$$ form a G.P. with common ratio, $$r = \sqrt{2}$$

It is given that, $$S_3 = ar^2 = 4$$

=> $$a (\sqrt{2})^2 = 4$$

=> $$a = \frac{4}{2} = 2$$

$$\therefore$$ $$n^{th}$$ term of G.P. = $$a (r^{n - 1})$$

= $$2 (\sqrt{2})^{n - 1}$$

=$$2^[{\frac{n+1}{2}}]$$

Given that the numbers are in G.P.

Let the common ratio be ‘r’, hence the series a,b,c,d,e can also be expressed as:

$$a , ar , ar^2 , ar^3 , ar^4$$

lcm(a,b) = lcm$$(a,ar) = ar$$

lcm(b,c) = lcm$$(ar,ar^2) = ar^2$$

lcm(c,d) = lcm$$(ar^2,ar^3) = ar^3$$

lcm(d,e) = lcm$$(ar^3,ar^4) = ar^4$$

$$\therefore \frac{1}{lcm(a,b)}+\frac{1}{lcm(b,c)}+\frac{1}{lcm(c,d)}+\frac{1}{lcm(d,e)}$$

= $$\frac{1}{ar} + \frac{1}{ar^2} + \frac{1}{ar^3} + \frac{1}{ar^4}$$

= $$\frac{1}{a} (\frac{1}{r} + \frac{1}{r^2} + \frac{1}{r^3} + \frac{1}{r^4})$$

To get max value of this, ‘a’ and ‘r’ should be minimum.

It is given that $$1 \leq a$$ => Minimum value of ‘a’ = 1

For the values in the series to be integers, the minimum common ratio, r = 2   ($$r \leq 1$$ won’t work here as it is an increasing GP)

Substituting values of 'a' and 'r' in the expression, we get :

Max value = $$\frac{1}{1} (\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4})$$

= $$\frac{8 + 4 + 2 + 1}{16} = \frac{15}{16}$$