The series is neither Arithmetic nor Geometric series. So it can be split into two series and then find the sum.
5 + 55 + 555 + ... + n terms  = 5 [1 + 11 + 111 + ...n terms]
= \(\frac { 5 }{ 9 } \) [9 + 99 + 999 +...+ n terms]
= \(\frac { 5 }{ 9 } \)[(10 - 1) + (100 - 1) + (1000 - 1) +...+ n terms]
= \(\frac { 5 }{ 9 } \)[10 + 100 + 1000 +...+ n terms)-n]
= \(\frac { 5 }{ 9 } \left[ \frac { 10\left( { 10 }^{ n }-1 \right) }{ \left( 10-1 \right) } -n \right] =\frac { 50\left( { 10 }^{ n }-1 \right) }{ 81 } =\frac { 5n }{ 9 } \)

If S₁, S₂ and S₃ are sum of first n, 2n and 3n terms of an A.P respectively then 
\({ S }_{ 1 }=\frac { n }{ 2 } \left[ 2a+\left( n-1 \right) d \right] ,{ S }_{ 2 }=\frac { 2n }{ 2 } \left[ 2a+\left( n-1 \right) d \right] ,{ S }_{ 3 }=\frac { 3n }{ 2 } \left[ 2a+\left( 3n-1 \right) d \right]\)
Consider, S₂ - S₁ = \(\frac { 2n }{ 2 } \left[ 2a+\left( n-1 \right) d \right] -\frac { n }{ 2 } \left[ 2a+\left( n-1 \right) d \right] \)
\(\frac { n }{ 2 } \left[ \left[ 4a+2\left( 2n-1 \right) d \right] -\left[ 2a+\left( n-1 \right) d \right] \right] \)
S₂ - S₁ = \(\frac { n }{ 2 } \left[ 2a+\left( 3n-1 \right) d \right] \)
3(S₁ - S₂) = \(\frac { n }{ 2 } \times \left[ 2a+\left( 3n-1 \right) d \right] \)
3(S₂ - S₁) = S₃

Given the 13^th term = 3 so, t₁₃ = a + 12d = 3..... (1)
Sum of first 13 terms = 234 gives \(\frac { 13 }{ 2 } \) [2a + 12d] = 234
2a + 12 = 36...(2)
Solving (1) and (2) we get , a = 33, d = \(\frac { -5 }{ 2 } \)
Therefore, common difference is \(\frac { -5 }{ 2 } \).
Sum of first 21 terms S₂₁ = \(\frac { 21 }{ 2 } \left[ 2\times 33+\left( 21-1 \right) \times \left( -\frac { 5 }{ 2 } \right) \right] =\frac { 21 }{ 2 }\)[66 - 50] = 168.

Let the first month earning be
t1 = Rs. 15,000
Increase per year d = Rs. 1,500
Expenses for first year =  Rs. 13,000
Expenses increase per year = Rs. 900
Saving for every year will be
(15000 - 13000), (16500 = 13900), (18000 - 14800),...
i.e. 2000, 2600, 3200, ...
Since t₂ - t₁ - t₃ - t₂, this sequence form an A.P
a = 2000, d = 2600 - 2000 = 600
Let the n^th years saving be Rs 20,000 then
a + (n - 1)d = t_n
2000 + (n - 1) (600) = 20,000
(n - 1) 600 = 20,000 - 2000
(n - 1) 600 = 18000
\(n-1=\frac{18000}{600}=30\)
n = 30 + 1 = 31
To save Rs. 20000, it takes 31 years.

Let the A.P. be t₁, t₂ , t₃, t₄,....
It is given that t₇ = -1 and t₁₆ = 17
a + (7 -1)d = -1 and a + (16 - 1)d = 17
a + 6d = -1 .....(1)
a + 15d = 17 ......(2)
Subtracting equation (1) from equation (2), we get 9d = 18 gives d = 2
putting d = 2 in equation (1), we get a + 12 = -1 so a = -13
Hence, General term t_n  = a + (n - 1)d
= -13 + (n -1) x 2 = 2n - 15 

Starting time from Chennai = 23.30 hrs
Travelling time = 11 hrs
Here we use modulo 24.
Reaching time = 23.30 + 11 (mod 24)
= 34.30 (mod 24)
= 10.30 (mod 24)
Since 11 = 0 x 24 + 11
It reaches London on Monday at 10.30 a.m
Chennai time = 4.30 hrs + London time
London time = Chennai time - 4.30 a.m
= 10.30 - 4.30 = 6 a.m
The flight will land at London Airport on Monday at 5 a.m.

First take 2⁵ ≡ 15 (mod 17)
(25)² ≡ 15² (mod 17)
≡ 4 (mod 17)
2¹⁰ ≡ 4 (mod 17)
(2¹⁰)⁴ ≡ 4⁴ (mod 17)  
2⁴⁰ ≡ 1 mod 17
(2⁴⁰)² ≡ 1² mod 17
2.2⁸⁰ ≡ 2 x 1 (mod 17)
2⁸¹ ≡ 2 mod 17
Remainder when 2⁸¹ is divided by 17 is 2.

Given 113400 = \({ p }_{ 1 }^{ { x }_{ 1 } }\times { p }_{ 2 }^{ { x }_{ 2 } }\times { p }_{ 3 }^{ { x }_{ 3 } }\times { p }_{ 4 }^{ { x }_{ 4 } }\)

The number 113400 can be factorized as 113400 = 2³ x 3⁴ x 5² x 7¹ where 2,3,5,7 are primes in ascending order.
Comparing 
\({ p }_{ 1 }^{ { x }_{ 1 } }\times { p }_{ 2 }^{ { x }_{ 2 } }\times { p }_{ 3 }^{ { x }_{ 3 } }\times { p }_{ 4 }^{ { x }_{ 4 } }\) = 2³ x 3⁴ x 5² x 7¹
P₁ = 2; P₂ = 3; P₃ = 5; P₄ = 7
x₁ = 3; x₂ = 4; x₃ = 2; x₄ = 1.

Using Euclid’s Division Algorithm, let us find the HCF of given numbers
210 = 55 x 3 + 45
55 = 45 x 1 + 10 
45 = 10 x 4 + 5
10 = 5 x 2 + 0
The remainder is zero.
So, the last divisor 5 is the Highest Common Factor (HCF) of 210 and 55.
Since, HCF is expressible in the form  55x - 325 = 5
gives 55x = 330
Hence x = 6

2 + 4 + 6 +...+ 80 = 2 (1 + 2 + 3 +..+ 40) = 2 x \(\frac { 40\times \left( 40+1 \right) }{ 2 } =1640\)

The general form of Geometric progression is a, ar, ar²,...
a = 256, ar = 256 x 0.5 = 128 , ar² = 256 x (0.5)² = 64
Therefore the required Geometric progression is 256,128, 64,....   

98 \(\equiv \) (x + 4) (mod 5)
98 - (x + 4) = 5n , for some integer n.
94 - x  = 5n
94- x is a multiple of 5
Therefore , the least positive value of x must be 4
Since 94 - 4 = 90  is the nearest multiple of 5 less than 94.

Since, 1 + 2 + 3 +...+ n = \(\frac { n\left( n+1 \right) }{ 2 } \), we have \(\frac { n\left( n+1 \right) }{ 2 } \) = 666
n² + n - 1332 = 0 gives (n + 37) (n - 36) = 0
So, n = -37 or n = 36
But n \(\neq \) -37 (Since n is a natural number); Hence n = 36

1+ 2 + 3 + .. + 50
Using , 1 + 2 + 3 + ...+ n = \(\frac { n\left( n+1 \right) }{ 2 } \)
1 + 2 + 3 + ... + 50 = \(\frac { 50\times \left( 50+1 \right) }{ 2 } \) = 1275 

Here a = 3,  r =  \(\frac { { t }_{ 2 } }{ { t }_{ 1 } } =\frac { 1 }{ 3 } \)
Sum of infinite terms = \(\frac { a }{ 1-r } =\frac { 3 }{ 1-\frac { 1 }{ 3 } } =\frac { 9 }{ 2 } \)

Common ratio = 4 > 1, sum of first 6 terms S₆ = 4095
Hence , S₆ = \(\frac { a\left( { r }^{ n }-1 \right) }{ r-1 } =4095\)
Since, r = 4,\(\frac { a\left( 4^{ 6 }-1 \right) }{ 4-1 } \) = 4095 gives a x \(\frac { 4095 }{ 3 } =4095\)
First term a = 3.

Total amount to repay = Rs. 65000
He pays Rs. 400 in the first installment and increasing the payment by Rs. 300 every month.
This form an A.P.
i.e., 400, 700, 1000, .....
a = 400,.d = 300
Sum upto n terms
\(\mathrm{S}_{\mathrm{n}} =\frac{n}{2}[2 a+(n-1) d] \)
\(65000 =\frac{n}{2}[2(400)+(n-1) 300] \)
\(65000 =\frac{n}{2} \times 2[400+(n-1) 150] \)
= n[400 + 150n - 150]
= n[150 n + 250]
65000 = 150n² + 250n
Divided by 50
1300 = 3n² + 5n
3n² + 5n - 1300 = 0
\(\mathrm{n} =\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \)
\(=\frac{-5 \pm \sqrt{5^{2}-4(3)(-1300)}}{2(3)} \)
\(=\frac{-5 \pm \sqrt{25+15600}}{6}=\frac{-5 \pm \sqrt{15625}}{6} \)
\(=\frac{-5 \pm 125}{6} \)
\(n =\frac{-5-125}{6}, \frac{-5+125}{6} \)
\(\mathrm{n} =\frac{-130}{6}, \frac{120}{6} \)
\(=\frac{-130}{6}, 20 \)
n cannot be negative
n = 20.
He will clear the loan by 20 months

Here the first term a = 8, common difference d = \(7\frac { 1 }{ 4 } \) -8 = -\(\frac { 3 }{ 4 } \).,
Sum of first n terms of an A.P S_n = \(\frac { n }{ 2 } \left[ 2a+\left( n-1 \right) d \right] \)
S₁₅ = \(\frac { 15 }{ 2 } \left[ 2\times 8+\left( 15-1 \right) \left( -\frac { 3 }{ 4 } \right) \right] \)
S₁₅ = \(\frac { 15 }{ 2 } \left[ 16-\frac { 21 }{ 2 } \right] =\frac { 165 }{ 4 } \)

Let the number of seats in the front row
t₁ = 20
Number of seats in the second row
t₂ = 20 + 2 = 22
Let the numbers of seats in the 30^th row be t₃₀. Since the additional seats in each row is 2.
t₁, t₂, ....t₃₀ form an A.P.
i.e., 20, 22, 24,... upto 30 terms
Here a = 20; d = 2
t_n = a + (n- 1) d form an A.P.
t₃₀ = 20 + (30- 1) (2) = 20 + 29 x 2
= 20 + 58 - 78
In the last row there are 78 seats.

We know that n^th term of an A.P. is
t_n = a + (n - 1)d
Given 9 times 9^th term = 15 times 15^th term
9 x t₉ = 15 x t₁₅
9[a + (9 - 1)d] = 15 [a + (15 - 1)d]
9(a + 8d) = 15 (a + 14d)
9a + 72d = 15a + 210 d
15a + 210 d - 9a - 72 d = 0
6a + 138 d = 0
6(a + 23 d) = 0
6[a + (24 - 1)d) = 0
6 x t₂₄ = 0
6 times 24^th term = 0

First term a = 3; common difference d = 6 - 3 = 3 ; last term l  = 111
We know that, n  = \(\left( \frac { l-a }{ d } \right) +1\)
n = \(\left( \frac { 111-3 }{ 3 } \right) +1\) = 37
Thus the A.P. contain 37 terms 

Let us associate the numbers 0, 1, 2, 3, 4, 5, 6 to represent the weekdays from Sunday to Saturday respectively.
Vani says today is Monday. So the number for Monday is 1. Since Vani’s birthday was 75 days ago, we have to subtract 75 from 1 and take the modulo 7, since a week contains 7 days.
–74 (mod 7) \(\equiv \)  –4 (mod 7) \(\equiv \) 7– 4 (mod 7) \(\equiv \) 3 (mod 7) 
(Since −74 - 3 = 77 is divisible by 7)
 Thus, 1 - 75 \(\equiv \) 3 ( mod 7)
The day for the number 3 is Wednesday.
Therefore, Vani’s birthday must be on Wednesday.

15 \(\equiv \) 3 (mod d) means 15 - 3 = kd, for some integer k,
12 = kd
gives d divides 12.
The divisors of 12 are 1,2,3,4,6,12. But d should be larger than 3 and so the possible values for d are 4, 6, 12.