MAH MBA CET Series and Progressions Questions

Let the first term be "a" and the common difference of the AP be "d".

So, 14th terms of AP will be a+13d.

a+13d = 202 ----equation (1)

19th term of AP will be a+18d.

a+18d = 277    --------equation (2)

solving both equations, we get a = 7 and d = 15

Therefore, first term = 7

8th term = a+7d = 7+7*15 = 112

So, ratio = 7:112 = 1:16

Given that the first term of GP is 16 and sum till infinity is 24

So, $$S_∞=\dfrac{a}{1-r}$$, where $$a$$ is the first term and $$r$$ is the common ratio.

$$S_∞=\dfrac{16}{1-r}=48$$

$$1-r=\dfrac{16}{48}=\dfrac{1}{3}$$

or, $$r=\dfrac{2}{3}$$

Now, we have to find sum till infinity of another GP whose first term is 76 and common ratio is $$r_1=\left(r-\dfrac{1}{6}\right)=\dfrac{2}{3}-\dfrac{1}{6}=\dfrac{1}{2}$$

$$S_∞=\dfrac{76}{1-\dfrac{1}{2}}$$

$$S_∞=152$$

Hence, the answer is 152

a, b, c, d and e are distinct.
Given that c is the arithmetic mean of a and b, there can be 2 cases. 
Case 1 : The sequence is a,c,b
Case 2: The sequence is b,c,a
Given that d is the arithmetic mean of b anc c, there can be 2 cases.
Case 1 : The sequence is b,d,c
Case 2: The sequence is c,d,b
Only 2 final cases are possible here.
Sequence 1 : a,e,c,d,b
Sequence 2 : b,d,c,e,a

Statement (i) is true as in both sequence 1 and sequence 2, the average of all terms is c.
Statement (ii) is true as in both sequence 1 and sequence 2, the average of d and e is c and average of a and b is c. c is not greater than c.
Statement (iii) is false as average of b and c, which is d is not greater than average of a and d in both sequences simultaneously.
Option A is the answer.

We are noticing that each term is obtained by dividing the previous term by 2.

So, the next term: $$\dfrac{1}{4}\ \times\dfrac{1}{2}=\dfrac{1}{8}$$

There is 1 ball on top, 3 balls in second layer, 6 balls in third layer,10 balls in fourth layer .......... and so on 
Total balls=8436
Sum of balls across all layers = S = 1+3+6+10+.........
S = 1+3+6+10+.......+$$t_n$$ 
S =     1+3+6+........+$$t_{n-1}\ +\ t_n$$
Subtracting both equations,
0 = 1+2+3+4+......... - $$t_n$$
$$t_n$$ = 1+2+3+4+.......... = $$\ \frac{\ n\left(n+1\right)}{2}$$
$$S_n=\Sigma\ t_n=\Sigma\ \ \frac{\ n\left(n+1\right)}{2}=\ \frac{\ 1}{2}\Sigma\ \left(n^2+n\right)$$
$$S_n=\ \frac{\ 1}{2}\left(\ \frac{\ n\left(n+1\right)\left(2n+1\right)}{6}+\ \frac{\ n\left(n+1\right)}{2}\right)=\ \frac{\ n\left(n+1\right)\left(n+2\right)}{6}=8436$$
n(n+1)(n+2)=50616
On solving, we get n=36.
Option B is the answer.

The Rabbit covers the distance in a geometric progression, with common ratio i.e 'r'= 1/5.

Since it travelled 12000 meters on the first day, the first term of the series is 12000

Now, we know that $$0<r<1$$ and in that case, the geometric progression is infinite. 

Hence, maximum distance it can cover = Sum of infinite GP= $$\dfrac{\ a}{1-r}$$, where a = 12000 and r = 1/5

So, combined maximum distance the rabbit can travel on all the days = $$\dfrac{12000}{1-\dfrac{1}{5}}\ =\ \dfrac{12000}{\dfrac{4}{5}}=12000\times\dfrac{5}{4}=15000$$ meters

The terms of series 1 is of the form 3n+2. The terms of series 2 is of the form 20+2m

3n+2 = 20+2m

when n = 6 and m = 1, the above equation satisfies. So, the first common term is 20.