The Sum of all terms of the arithmetic progression having 10th terms except for the 1st term, is 99, and except for the 6th term , 89. Find the 3rd terms of the progression if the sum of the 1st and the 5th term is equal to 10.

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

According to question,
First Two Digit Number, which is divided by 7 and give the remainder 3, will be 10.
Last Two Digit Number, which is divided by 7 and give the remainder 3, will be 94.
The common difference between two consecutive numbers will be 7.
This series will be like ?10, 17, 21 ,............................ 94.
Here First Number a = 10, Last number t_n = 94, Common Difference d = 7 ;
Using the Formula for Last Number t_n = a + (n - 1) x d;
? a + (n - 1) x d = t_n
Put the value of a , d and t_n in above equation.
? 10 + (n - 1) x 7 = 94
? 10 + 7n - 7 = 94
? 7n + 3 = 94
? 7n = 94 - 3 = 91
? n = 91/7
? n = 13
Using the formula for the sum of Arithmetic Progression.
S_n = n/2 [ First Number + Last Number ] ;
Put the value of n , First Number and Last Number, we will get,
S_n = 13/2 [ 10 + 94 ]
S_n = 104 x 13/2
S_n = 52 x 13
S_n = 676

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

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

Let the number of rice bowls be a, the number of juice bowls be b, and the number of meat bowls be c.
According to question,
a + b + c = 65........................(1)
The total number of guests = 2a
The total number of guests = 3b
The total number of guests = 4c
So the total number of guests will be same in the party.
2a = 3b = 4c..........................(2)
As per Equation (2)
b = 2a/3................................(3)
c = 2a/4 = a/2......................(4)
Now put the value of b and c from Equation (3), (4) in Equation (1),
a + 2a/3 + a/2 = 65
(6a + 4a + 3a)/6 = 65
13a = 65 x 6
a = 5 x 6 = 30
Put the value of a in equation (3) and (4) in order to get the value of b and c,
b = 2 x 30/3 = 2 x 10 = 20
c = 30/2 = 15

The Total number of Guests = 2a = 3b = 4c = 60

Let us assume Ram has R rupees and Mohan has M rupees.
According to question,
If Ram gives 30 rupees to Mohan, then
Ram has left money = R - 30 and Mohan has money = M + 30
Then Mohan will have twice the money left with Ram,
M + 30 = 2(R - 30)
M + 30 = 2R - 60
2R - M = 90................................(1)
Again According to question,
if Mohan gives 10 rupees to Ram, then
Mohan has left the money = M - 10 and Ram has the money = R + 10
Then According to question,
Ram will have thrice as much as is left with Mohan,
R + 10 = 3 (M - 10 )
? R + 10 = 3M - 30
? 3M - R = 10 + 30
? 3M - R = 40..................................(2)
After Multiplying 2 with Equation (2) , add with the equation (1),
6M - 2R + 2R - M = 80 + 90
? 6M - M = 170
? 5M = 170
? M = 170/5
? M = 34
Put the value of M in equation (1), we will get
? 2R - 34 = 90
? 2R = 90 + 34
? 2R = 124
? R = 124/2
? R = 62