Some persons can do a piece of work in 12 days. Two times the number of such persons will do half of that work in :

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.

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

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.

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