what is the sum of all the two-digits numbers which when divided by 7 gives a remainder of 3?

According to given question,
First day 1 rs, second day 2 rs, third day 4 rs.....
1 , 2 , 4............ and so on
The given series is in geometric progression (GP).
1, 2¹, 2²,.............
Sum of first n terms in a geometric progression(GP)
S_n = a (rⁿ ? 1) /(r ?1)
if r > 1 and where a= the first term, r = common ratio,n = number of terms
As per the given question a = 1 , r = 2 and n = 20;
S_n = 1(2²⁰ ? 1) /(2 ?1)
S_n = (2²⁰ ? 1) /1
S_n = (2²⁰ ? 1)
S_n = 2²⁰ ? 1

Let us assume the second number is a and the difference between consecutive numbers is d.
According to Arithmetic progression,
First number = a - d
Second number = a
Third number = a + d
According to question,
Sum of the all three numbers = 15
a - d + a + a + d = 15
3a = 15
a = 5
Again according to given question,
sum of square of the 3 numbers = 83
(a - d) ² + a ² + (a + d) ² = 83
apply the algebra formula
a ² + d ² - 2ad + a ² + a ² + d ² + 2ad = 83
3a ² + 2d ² = 83
Put the value of a in above equation.
3 x 5 ² + 2d² = 83
3 x 25 + 2d² = 83
75 + 2d² = 83
2d ² = 83 - 75
2d ² = 8
d ² = 8/2
d ² = 4
d = 2
Put the value of a and d in below equation.
First number = a - d = 5 - 2 = 3
Second number = a = 5
Third number = a + d = 5 + 2 = 7
The smallest number is 3.

Let us assume the first installment is a and difference between two consecutive installments is d.
According to question,
Sum of 40 Installments = 3600
Apply the formula Sum of first n terms in an Arithmetic Progression = S_n = n/2[ 2a + (n?1)d ]
where a = the first term, d = common difference, n = number of terms.
n/2[ 2a + (n?1)d ] = 3600
Put the value of a, n and d from question,
40/2[ 2a + (40 ?1)d ] = 3600
20[ 2a + 39d ] = 3600
[ 2a + 39d ] = 3600/20 = 180
2a + 39d = 180...........................(1)

Again according to given question,
After paying the 30 installments the unpaid amount = 1/3(total unpaid amount) = 3600 x 1/3
After paying the 30 installments the unpaid amount = 1200
So After paying the 30 installments the paid amount = 3600 - 1200 = 2400
Sum of 30 installments = 2400
30/2[ 2a + (30 ?1)d ] = 2400
[ 2a + (30 ?1)d ] = 2400 x 2/30
2a + 29d = 80 x 2 = 160
2a + 29d = 160..........................(2)
Subtract the Eq. (2) from Eq. (1), we will get
2a + 39d - 2a - 29d = 180 - 160
10d = 20
d = 2
Put the value of d in Equation (1), we will get
2a + 39 x 2 = 180
2a = 180 - 78
a = 102/2
a = 51
The value of first installment = a = 51

According to the question,
Every day adding 1 rs extra to previous day.
Let us assume after n day, total saving will become perfect square.
1 + 2 + 3 + 4 + 5 + 6 +.......................+ t_n.
Apply the algebra A.P formula,
sum of total rupees after n days = n(n+1)/2
Hit and trail method, Put the value of n = 2 , 3 , 4, 5 .... and so on to get the perfect square.
If n = 2
n(n+1)/2 = 2 x 3 / 2 = 3 which is not perfect Square.
If n = 3
n(n+1)/2 = 3 x 4 / 2 = 6 which is not perfect Square.
If n = 4
n(n+1)/2 = 4 x 5 / 2 = 10 which is not perfect Square.
If n = 5
n(n+1)/2 = 5 x 6 / 2 = 15 which is not perfect Square.
If n = 6
n(n+1)/2 = 6 x 7 / 2 = 21 which is not perfect Square.
If n = 7
n(n+1)/2 = 7 x 8 / 2 = 28 which is not perfect Square.
If n = 8
n(n+1)/2 = 8 x 9 / 2 = 36 which is perfect Square.

n(n+1)/2 should be a perfect square . The first value of n when this occurs would be for n = 8. thus , on the 8th of March of the required condition would come true.

First 3-digit number divisible by 6 is 102 and last 3-digit number divisible by 6 is 996.
Difference between two consecutive numbers divisible by 6 is 6.
So 3-digit numbers divisible by 6 are 102,108,114, ......., 996.
This is an Arithmetic Progression in which a = 102, d = 6 and l = 996.
where a = First Number , l = Last Number and d = difference of two consecutive numbers.
Let the number of terms be n. So Last term = t_n
Then t_n = 996
Use the formula for n terms of arithmetic progression.
? a + ( n - 1) x d = 996
? 102 + (n - 1) x 6 = 996
? 6(n - 1) = 894
? (n - 1) = 149
? n = 150
? Numbers of terms = 150

let us assume a be the first term and d the common difference.
According to Question,
m^th term of an Arithmetic Progression (AP) is 1/n
Use the Arithmetic Progression formula for m^th term,
? a + ( m - 1 ) x d = t_m
? a + ( m - 1 ) x d = 1/n
? a + md - d = 1/n .......................(1)
Use the Arithmetic Progression formula for n^th term,
? a + ( n - 1 ) x d = t_n
? a + ( n - 1 ) x d = 1/m
? a + nd - d = 1/m.........................(2)
Subtracts Equation (1) from Equation (2) , we will get
? a + nd - d - (a + md - d) = 1/m - 1/n
? a + nd - d - (a + md - d) = 1/m - 1/n
? a + nd - d - a - md + d = 1/m - 1/n
? nd - md = 1/m - 1/n
? d(n - m) = 1/m - 1/n
? d(n - m) = (n - m)/mn
? d = 1/mn.......................................(3)
Put the value of d in Equation (1), we will get,
? a + md - d = 1/n
? a + ( m x 1/mn ) - 1/mn = 1/n
? a + 1/n - 1/mn = 1/n
? a = 1/n - 1/n + 1/mn
? a = 1/mn.....................................(4)
Now according to question,
Sum of mn terms = mn/2(2a + ( mn - 1) x d)
S_mn = mn/2(2a + ( mn - 1) x d)
Put the value of a and d in above equation, we will get,
S_mn = mn/2 [ 2 x 1/mn + ( mn - 1) x 1/mn ]
S_mn = mn/2 [2/mn + 1 - 1/mn ]
S_mn = mn/2 [ 1 + 1/mn ]
S_mn = mn/2 [ ( mn + 1 )/mn ]
S_mn = mn/2 (1 + mn)/mn
S_mn = 1/2 (1 + mn)
S_mn = (1 + mn)/2 = (mn + 1)/2

Let us assume the first term is a and common difference is d.
Formula for n^th term of Arithmetic Progression = a + ( n - 1 ) x d
So first term t₁ = a + ( n - 1 ) x d
t₁ = a + (1 - 1) x d = a + 0 x d = a
Fifth term t₅ = a + ( n - 1 ) x d = a + (5 - 1) x d = a + 4d
According to question,
Sum of the first term and the fifth term = 10
t₁ + t₅ = 10
? a + a + 4d = 10
? 2a + 4d = 10
? a + 2d = 5.............................(1)

Formula of sum of n terms for Arithmetic Progression,
S_n = n/2[ 2a + ( n - 1 ) x d ]
Put the value of n = 10, Since total number of terms is 10 .
? S₁₀ = 10/2[ 2a + ( 10 - 1 ) x d ]
? S₁₀ = 5[ 2a + 9 x d ] = 10a + 45d
Again According to question,
The sum of all terms of the Arithmetic Progression, except for the 1st term = 99
S₁₀ - first term = 99
? 10a + 45d - a = 99
? 9a +45d = 99
? a + 5d = 11.................................(2)
Subtracts Equation (1) from Equation (2), we will get,
? a + 5d - a - 2d = 11 - 5
? 3d = 6
? d = 2
Put the value of d in equation (1), we will get
? a + 2 x 2 = 5
? a + 4 = 5
? a = 5 - 4
? a = 1
Since t_n = a + ( n - 1) x d
find the third term by putting the value of a , n and d. we will get,
t₃ = 1 + ( 3 - 1) x 2 = 1 + 2 x 2 = 1 + 4 = 5
So 3rd term of Arithmetic Progression is 5.

Let us assume the number is n.
According to question,
If 24 is subtracted, it becomes 4/7 of the number.
n - 24 = n x 4/7
? n - 4n/7 = 24
? (7n - 4n)/7 = 24
? 3n/7 = 24
? 3n = 24 x 7
? n = 24 x 7/3
? n = 8 x 7
? n = 56
Sum of the digits of the number = 5 + 6 = 11

Let us assume the number is a.
Given that 7 ¹/3 = 22/3
According to question,
a x 1/2 + a x 1/5 = a x 1/3 + 22/3
? a/2 + a/5 = a/3 + 22/3
? a/2 + a/5 - a/3 = 22/3
? (15a + 6a - 10a)/30 = 22/3
? (15a + 6a - 10a) = 30 x 22/3
? 11a = 10 x 22
? a = 10 x 2
? a = 20

Let us assume the salary of driver be ? R
Then Tips of week = R x 5/4
According to question,
his income during one week = Salary + Tips
? Total Income during one week = R + (5R/4)
? Total Income during one week = (4R + 5R)/4 = 9R/4

Required Fraction = Tips in a week/Total Income in a week
? Required Fraction = 5R/4 / 9R/4
? Required Fraction = 5R/4 x 4/9R
? Required Fraction = 5/9