A man arranges to pay off a debt of ?3600 in 40 annual installments which form an Arithmetic Progression (A.P). When 30 of the installments are paid, he dies leaving one-third of the debt unpaid . Find the value of the first installment.

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 the first number is a and common difference is d.
According to question,
4th term of A.P = 37
a + ( n - 1 ) x d = 37
Put the value of a , n and d, we will get,
a + (4 - 1 ) x d = 37
a + 3d = 37..................(1)
sixth term is 12 more than the fourth term,
6th term = 12 + 4th term
a + ( n - 1 ) x d = 12 + 37
a + ( 6- 1 ) x d = 39
a + 5d = 39................(2)
subtract the equation (1) from (2)
a + 5d - a - 3d = 39 - 37
5d - 3d= 2
2d = 2
d = 1
Put the value of d in equation (1), we will get
a + 3 x 1 = 37
a = 37 - 3
a = 34

Second term = a + (n - 1) x d = 34 + (2 - 1) x 1 = 34 + 1 = 35
Six term = a + (n - 1) x d = 34 + (6 - 1) x 1 = 34 + 5 = 39
Sum of Second and Six term = 35 + 39 = 74
Sum of Second and Six term = 74
Answer is 74.

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.

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

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

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.