Answer: Option B. -> 42
:
B
$\frac{n(n+1)}{2}-\frac{1+2+3}{2}=900:\frac{n(n+1)}{2}=903;$
n(n+1) = 1806, n = 42.

Answer: Option B. -> 49{109(10n−1)−n}
:
B
Reverse Gear Approach
Assume n= 1, then $S_1$=4
Now put n=1in options.Eliminate those options , where you are not getting 4.
Only option (b) will give 4.

Answer: Option B. -> 25
:
B
We need to use reverse of Euler's theorem as the Euler's number for 29=28.
And the question reduces to $\left(\frac{7^{27}}{29}\right)$ (Rem).
The right answer multiplied by 7, should give a remainder of 1 when divided by 29, as $\left(\frac{7^{28}}{29}\right)$ (Rem) = 1.
Thus, the answer is option (b).
Verification -
25 or $(-4)\times 7$= -28 $⟹$ remainder of 1.

:
Let the number be x.
According to the statement,
$290-\frac{x}{6}=\frac{x}{2}$
or $\frac{x}{2}+\frac{x}{6}=290$
or $\frac{4x}{6}=290$
or x = 435.
Therefore, the number is 435.

Answer: Option A. -> 120m
:
A
The last person covered 120.71 meters.
It is given that the platoon and the last person moved with uniform speed. Also, they both moved for the identical amount of time. Hence, the ratio of the distance they covered - while person moving forward and backward - are equal. Let's assume that when the last person reached the first person, the platoon moved X meters forward.
Thus, while moving forward the last person moved (50+X) meters whereas the platoon moved X meters.
Similarly, while moving back the last person moved [50-(50-X)] = X meters whereas the platoon moved (50-X) meters.
Now, as the ratios are equal,
$\frac{[50+X]}{X}=\frac{X}{[50-X]}$
$[50+X]\times [50-X]=X\times X$
Solving, X=35.355 meters
Thus, total distance covered by the last person
$=[50+X]+X$
$=2\times X+50$
$=2\times \left[35.355\right]+50$
=120.71 metres.
=120 m (Approximately)

Answer: Option D. -> None of these
:
D
Let x=y+1so, $x^3-y^3=1+3y(y+1)$clearly this is not divisible by either 2 or 3 so m = 0 and n = 0 so m+n=0.
Hence option (d)

Answer: Option A. -> 2T1
:
A
Suppose the distance between the two points is 'd'.
When the persons meet for the first time, total distance covered = d
When persons meet for the second time, total distance covered = d + 2d
Distance = d, time taken = T1; distance = 2d, time taken = 2T1 = T2
Hence Option (a)
Alternative Approach: Using Assumption
You can also solve this using numbers. If the distance is 100 km, and the speed is 50 kmph for each person, they will meet after $\frac{1}{2}$ an hour. They will reach the opposite corners in 1 hour and will meet again in $1\frac{1}{2}$ hours; thus T₂=1 hour = 2T1. Option (a)

Answer: Option B. -> 32
:
B
Following the pattern
A${}_1$=$\pi$
A${}_2$=4$\pi$
A${}_3$=9$\pi$
A${}_4$=16$\pi$
Hencer = 1,2,3 4
Thus
C${}_1$= 2$\pi$
C${}_2$= 4$\pi$
C${}_3$= 6$\pi$
C${}_4$= 8$\pi$
Ratio = 1.5

Answer: Option D. -> 800
:
D
Consider the first round: - All the numbers which leave a remainder of 1 when divided by 15 will be marked. (1, 16, 31, 46 ... 991)
Consider the second round: - The first number to be marked is 991 + 15 - 1000 = 6. Thereafter all the numbers which leave a remainder 6 when divided by 15 will be marked. (6, 21, 36, 51 ... 996)
Consider the third round: - The first number to be marked is 996 + 15 - 1000 = 11. And all the numbers which leave a remainder 11 when divided by 15 will be marked. (11, 26, 41, 56 ... 986)
The first number to be marked in the fourth round is 986 + 15 - 1000 = 1. Now the cycle will repeat.
If we see the numbers which are getting marked are: 1, 6, 11, 16, 21, 26, 31, 36 ... 996
i.e. all the numbers which divided by 5 leave a remainder 1. So there are $\frac{1000}{5}=200$ numbers. So, 1000 - 200 = 800 numbers remain unmarked.

Answer: Option B. -> na1+a2n√a1+√an+1
:
B
Let the AP be 1,2,3,4,5
At n=1, the AP = 1,2
$a_1=1$ and $a_2=2$
The value of the expression $=\frac{3}{2.4}$
Only option (b) gives this value