Linear Equation
 The Fourth term of an Arithmetic Progression is 37 and the Sixth term is 12 more than the Fourth term. What is the sum of the Second and Eight terms?

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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)
Solve the equation and get the answer.Correct Option: C
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.
 How many 3digits numbers are completely divisible by 6?

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First 3digit number divisible by 6 is 102 and last 3digit number divisible by 6 is 996.
Difference between two consecutive numbers divisible by 6 is 6.
So 3digit 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 term of Arithmetic Progression.Correct Option: B
First 3digit number divisible by 6 is 102 and last 3digit number divisible by 6 is 996.
Difference between two consecutive numbers divisible by 6 is 6.
So 3digit 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
 On March 1st 2016 , sherry saved ₹ 1. Everyday starting from March 2nd 2016, he save ₹1 more than the previous day . Find the first date after March 1st 2016 at the end of which his total savings will be a perfect square.

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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.Correct Option: D
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.
 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 onethird of the debt unpaid . Find the value of the first installment.

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Let us assume the first installment is a and difference between two consecutive installments is d.
According to question,
Sum of 40 Installments = 3600
S_{40} = 40/2[ 2a + (40 ?1)d ] = 3600
Again according to given question,
Sum of 30 installments = 2400
S_{30} = 30/2[ 2a + (30 ?1)d ] = 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.Correct Option: C
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
 A number 15 is divided into 3 parts which are in Arithmetic Progression (A.P) and the sum of their squares is 83. What will be the smallest number?

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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
Again according to given question,
sum of square of the 3 numbers = 83
(a  d)^{ 2} + a^{ 2} + (a + d) ^{2} = 83
Solve the equation.Correct Option: B
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} + 2d^{2} = 83
3 x 25 + 2d^{2} = 83
75 + 2d^{2} = 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.