MAH MBA CET Series and Progressions Questions

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 terms form an AP with a common difference of -5.
Nth term of an AP = A + (N-1)D where A is the first term and D is the common difference.
Thus, 15th term will be 20 + (15-1)*(-5) = -50.

(1 x 2) + 1 = 3

(3 x 3) + 1 = 10

(10 x 2) + 1 = 21

(21 x 3) + 1 = 64

(64 x 2) + 1 = 129

(129 x 3) + 1 = 388

(388 x 2) + 1 = 777

Thus, the odd one out is 356

Hence, option E is the correct choice.

Here we can see that the common difference is 5 and he first term is 20.

We know that the formula of the th term is a+(n-1)d = 130

20+(n-1)5 = 130

n-1 = 22

n = 23

Nth term of an AP = A + (N-1)D where A is the first term and D is the common difference.
8th term = >A + 7D = 39.
12th term=> A + 11D = 59.
Subtracting the equations, we get 4D = 20 ==> D = 5.
Substituting this value in any of the equation gives us the 1st term i.e. A = 4.

Arithmetic mean = Sum/2 = 15/2.
Geometric mean = 0.8 * Arithmetic mean = 0.8 * 15/2 = 6.
Product of the 2 numbers will be square of Geometric mean = 36.
Let us assume the numbers to be x and y. 
x + y = 15 and xy = 36. Substituting value of x as 15 - y, we get:
y(15 - y) = 36 ==> $$y^2-15y\ +\ 36\ =\ 0$$.
y = 3 or 12.
So, the two numbers are 12 and 3.