Basically, we see that every day the number of particles increases by 50.
We need to reach 1000 particles from 100 particles on day 1.
Or, we need to increase by 900 particles.
At the rate of 50 particles per day, it takes 18 days to increase by 900 particles.
So on the 19^th day the number of articles will be 1000.

Given that, x, y, z are in arithmetic progression.
Also given that xyz = 5(x + y + z).
xyz = 5(3y). Since x, y, z are in arithmetic progression.
xz = 15.
The possible combinations for x and z are (3, 5) (1, 15).
 z – x > 4 (Since y – x > 2).
Therefore,
x = 1, y = 8, z = 15
Hence, z – x = 14.

We are given a generic formula in ‘n’ for the sum of first n terms, with alternating signs.
S_n = x₁ – x₂ + x₃ – x₄ + ….. + (-1)ⁿ⁺¹ x_n = n² + 2n = n(n + 2)
(Observe that the even terms have a negative sign, while the odd terms have a positive sign)
S₅₀ = 50 (50 + 2) = 50 (52) = 50 (2) (26) = 2600
S₄₉ = 49 (49 + 2) = 49 (51) = 2499
S₄₈ = 48 (48 + 2) = 48 (50) = (24) (2) (50) = 2400
S₅₀ = S₄₉ – x₅₀
x₅₀ = S₄₉ – S₅₀ = 2499 – 2600 = -101
S₄₉ = S₄₈ + x₄₉
x₄₉ = S₄₉ – S₄₈ = 2499 – 2400 = 99
x₄₉ + x₅₀ = 99 – 101 = -2
Alternate Solution:
Observe that n² + 2n is an increasing function for all natural numbers.
That is, if x > y; x² + 2x > y² + 2y.
S_n = n² + 2n = n(n + 2) = n² + 2n
S_{n – 1} = (n – 1)(n +1) = n² – 1
S_n – S_n-1 = n² + 2n – (n² – 1) = 2n + 1
So adding the n^th term to the sequence increases the sequence by 2n + 1
The even terms of the sequence are subtracted, but then the sum of the terms in the sequence still increases, this means that the n^th term is negative if n is even and n^th term is positive if n is odd, but the magnitude of the n^th term is is always (2n +1)
So, x_n = (-1)ⁿ⁺¹ (2n + 1)
x₅₀ = (-1)⁵⁰⁺¹ (2(50) + 1) = -101
x₄₉ = (-1)⁴⁹⁺¹ (2(49) + 1) = 99
x₄₉ + x₅₀ = 99 – 101 = -2

Let us take 5th kid has 2 toffees
Because we know that after 5th toffee his stock exhausts.
So only if the 5th kid has 2 toffees, he can give away half of it and 1 extra = 0
Then for 4th kid, (2+1) × 2 = 6 (Since we are moving in reversing order)
3rd = 14 and 2nd = 30 and 1st kid = 62

Given that a₁ – a₂ + a₃ – a₄ + …….. +(-1) ^n-1 a_n = n
So, a₁ = 1
a₁ – a₂ = 2 => a₂ = 1-2 = -1
a₁ – a₂ + a₃ = 3 => a₃ = 1
So, the series proceeds as 1, -1, 1, -1, …
Then a₅₁ + a₅₂ + …….. + a₁₀₂₃ = 1 + (-1) + …. + 1
a₅₁ + a₅₂ + …….. + a₁₀₂₃ = 1

First series – 15, 19, 23, 27, ……, 415 -> Common difference = 4
Second series – 14, 19, 24, 29, ……, 464 -> Common difference = 5
Common terms in both sequences = 19, 39, 59, …., (Common difference = LCM (4,5) = 20)
Now, observing the new series
19, 39, 59, …., = (20 – 1), (40 -1), (60 -1), …., (400-1) (There is no room for 419, as the first series ends at 415)
399 = 400 – 1 = 20 x 20 -1
Hence, the number of common terms in the two sequences = 20

Given series – (2n+1) + (2n+3) + (2n+5) + … + (2n+47) = 5280
Isolate 2n terms on one side
(2n + 2n +…. + 2n) + (1 +3 + 5 +…. + 47) = 5280
Odd numbers from 1 to 47 are added in the above series.
Number of terms from 1 to 47 = 24 terms
Therefore, the number of 2n terms = 24
For computing the value of (1 +3 + 5 +…. + 47),
We know 1 = 1², 1 +3 = 2², 1 +3 +5 = 3² and so on
So, (1 +3 + 5 +…. + 47) = 24²
So, 2n x 24 + 24² = 5280
2n x 24 = 5280 – 24²
2n = 220 – 24
n = 110 – 12
n = 98
So, value of 1+2+3+…+98 = = = 
Value of 1+2+3+…+98 = 4851

Let n = 1,
a₁ = 3(2¹⁺¹ -2) = 3(4-2) = 6 = 3 x 2¹
a₂ = 3(2²⁺¹ -2) = 3(8-2) = 18, a₂ = 18 – 6 = 12 = 3 x 2²
a₃ = 3(2³⁺¹ -2) = 3(14) = 42 a₃ = 42 – 12 – 6 = 42-18 =24 = 3 x 2³
So, a₁₁ = 3 x 2¹¹ = 3 x 2048
a₁₁ = 6144

It is given that a₁,a₂,……,a₅₂ are positive integers such that a₁ < a₂ < ……. < a₅₂ and it is also given that their arithmetic mean is one less than the arithmetic mean of a₂,……,a₅₂. Let us consider that a₂,……,a₅₂ have the arithmetic mean of M and from a₁ to a₅₂ the arithmetic mean is M – 1.
If we add a₁ to a₂,……,a₅₂, the average falls down by 1 or the total falls by 52. So, a₁ = M – 52

In conventional way if sum of a₂ + a₃ + …. + a₅₂ = 51M. Sum when a₁ is added = 51M + a₁. Average of a₁,a₂,……,a₅₂ = M – 1.
51M+a152 = M – 1
51M + a₁ = 52M – 52
a₁ = M – 52
We have to find the maximum value of a₁ and this value can be maximum when average or mean M tends to be maximum. We know that all the numbers are distinct i.e from a₂,……,a₅₂ there are 51 numbers. The maximum possible average of 51 numbers if the largest number is 100.
The numbers are i.e 50,51,…..,100. Average = 100+502100+502. Hence, the maximum possible average is 75.
M_max = 75
A_max = 75 – 52
A_max = 23
The largest possible value of a₁ is 23

Let t₁, t₂,…..be real numbers such that t₁+ t₂ +… + t_n = 2n² + 9n + 13
If t_k = 103 we have to find the value of k.
Let us assume this as sum to n terms i.e.
S_n = t₁+ t₂ +… + t_n = 2n² + 9n + 13
Given an expression in terms of n we can also take sum to n-1 terms i.e.
S_n-1 = t₁+ t₂ +… + t_n-1 = 2(n-1)² + 9(n-1) +13
This S_n and S_n-1 can be subtracted one from the another such that,
S_n = t₁+ t₂ +… + t_n = 2n² + 9n + 13
S_n-1 = t₁+ t₂ +… + t_n-1 = 2(n-1)² + 9(n-1) +13
S_n – S_n-1 = t_n = 2n² + 9n + 13 – 2(n-1)² – 9(n-1) – 13
S_n – S_n-1 = t_n = 2n² + 9n + 13 – 2(n-1)² – 9n + 9 – 13
S_n – S_n-1 = t_n = 2[n² – (n-1)²] + 9
S_n – S_n-1 = t_n = 2[n² – n² + 2n – 1] + 9

t_n = 2[2n – 1] + 9
t_n = 4n – 2 + 9
t_n = 4n + 7
So t_k = 4k + 7 = 103
k = 103−74103−74
k = 24
The 24^th term is 103

Here we have to find the value of the sum 7 × 11 + 11 × 15 + 15 × 19 + ….. + 95 × 99
T_n = (4n + 3) (4n + 7)
Since increment of 4 takes place in every value , 4n is the term to be used
T₁ = (4 + 3) (4 + 7) = 7 × 11
T₂ = 11 × 15
T₃ = 15 × 19 and so on
Expanding T_n = (4n + 3) (4n + 7)
We will get 16n² + 12n + 28n +21
16n² + 40n +21
Σ16n² + 40 Σn + 21 Σ1
16n(n+1)(2n+1)6 + 40n(n+1)2 + 21 × n
Simplifying this we can take n out
n[8n(n+1)(2n+1)3 + 20(n+1) + 21]
From the options we can take 80707 and check whether it is divisible by 23
80707238070723 = 3509
⟹ option a) 80707 is a multiple of 23
⟹ option b) 80751 is 80707 + 44 so this doesn’t work
⟹ option c) 80730 is 80707 + 23 so this is also a multiple of 23
⟹ option d) 80773 is 80707 + 66 so this also doesn’t work
So let us check out with option a and c so we have to substitute and simplify and find
n[8n(n+1)(2n+1)3 + 20(n+1) + 21]
n3[8(n+1)(2n+1) + 60(n+1) + 63]
From the answer choices we have only two possibilities left out i.e. 23 ⨯ 3509 or 23 ⨯ 3510
⟹ 233233[8(23 + 1)(2(23) + 1) + 60(23 + 1) + 63]
⟹ 233233[8(24)(47) + 60 (24) + 63]
⟹ 233233[8(24)(47) + 60 (24) + 63]
⟹ 23[8(8)(47) + 60 (8) + 21]
⟹ 80707
Hence the value of the sum 7 × 11 + 11 × 15 + 15 × 19 + ….. + 95 × 99 is 80707