## What is next terms in the series ?

Let us solve next number in sequence puzzle
15,29,56,108,208,????
Here 1st term is 15, To find out 2nd term if we multiply the 1st term With 2 and subtract 1 from it, i.e (15 *2) -1  = 29
To find 2nd term we have to multiply the previous term with 2 and subtract 2 (double the previous time) from it
I.e 29*2 - 2 = 56
To find 3rd term we have to multiply the previous term with 4   and subtract 2  (double the previous time) from it
I.e 56*2 - 4 = 108
To find 4th term we have to multiply the previous term with 8 (double the previous time) and subtract 2 from it
I.e 108*2 - 8 = 208
Similarly 5th term can be calculated as
multiply the previous term with 2 and subtract  subtract 2 (double the previous time) from it
I.e 208*2 - 16 = 400
Next term 400*2 - 32 = 768

Next term  768*2 - 64 = 1472
Next term  1472*2 - 128 = 2816
Or we can write it in another form as follows
15+(15–1) = 29
29+(29–2) = 56
56+(56–4) = 108
108+(108–8) = 208
208+(208–16) = 400
400+(400–32) = 768

768+(768–64= 1472

1472+(1472-128) =2816

And similarly this trend follows

## What’s the next number in the sequence 0, 2, 6, 12?

We can get the answer in two  ways

### 1st way

1 * (1–1) = 0 ( 1st term)

2 * (2–1) = 2 (2nd term)

3 * (3–1) = 6 (3rd term)

4 * (4–1) = 12 (4th term)

5 * (5–1) = 20 (5th term)

Next term must be  20
Similarly
6 * (6–1) = 30  (6th term)

7 * (7–1) = 42  (7th term)

### 2nd Way

1^2 –1 = 0

2^2 – 2 = 2

3^2 – 3 = 6

4^2 – 4 = 12

5^2 – 5 = 20
Next term must be  20
Similarly

6^2 – 6 = 30

7^2 – 7 = 42

## Can you solve this  Series: 30, 31, 28, 33, 26, 35, … What number should come next?

We can split this sequence into two series by picking alternate numbers from the given sequence.
30, 31, 28, 33, 26, 35, …
1st  one is 30,  28, 26, … and 2nd one is  31, 33,  35, …
Now after careful analysing the 1st sequence ,it next terms should be  with decrease of  2 ,we get other terms as follows
30,  28, 26, 24,22,20 …
And after careful analysing the 2nd  sequence ,it next terms should be  with increase of  2 ,we get other terms as follows
31, 33,  35, ,37,39,41…

26-2 = 24.
So the text terms should be 30, 31, 28, 33, 26, 35, 24,37,22

24 is the next number in the required sequence

____________________________________

## FINAL WORDS

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## HOW TO ADD 1 TO 1000 NUMBERS VERY EASILY

If you want to add 1 to 1000 then follow these two steps

2 Multiply last number and it's next numbers then divide by 2 . And thats it  .

अगर आप 1 से लेकर किसी भी अंतिम संख्या को जोड़ना चाहते हो तो सबसे पहले अंतिम संख्या में एक जोड़ दें फिर अंतिम संख्या और उससे आगे बाली संख्या को गुणा करके 2 से भाग दे दें ,आप इस तरह से कितनी भी शंख्ययाओं का जोड़ एक दम से कर सकते हो

### उदाहरण

If we have to add 1st 50 numbers then  Multiply 50 with 51 and divide the product with 2 and it is the sum of 1st 50 natural numbers

अगर हम 1 से 50 को जोड़ना चाहें तो 50 और 51 गुना करके 2 से भाग देंगे
1 +2+3+4+5+6+7+……+50 =(50×51)÷2 = 25×51=1275
अगर हम 1 से 100 को जोड़ना चाहें तो 100 और 101 गुना करके 2 से भाग देंगे
1+2+3+4+5+…….+100=( 100×101) ÷2 =50×101=5050
अगर हम 1 से 1000 को जोड़ना चाहें तो 1000 और 1001 गुना करके 2 से भाग देंगे
1+2+3+4+5+…….+1000 =(1000×1001)÷2 = 500×1001= 500500
अगर हम 1 से 2000 को जोड़ना चाहें तो 2000 और 2001 गुना करके 2 से भाग देंगे
1+2+3+4+5+…….+2000=( 2000×2001)÷2 =1000×2001 = 2001000 Share:

## HOW TO FIND SHORTEST DISTANCE BETWEEN TWO LINES

### SHORTEST DISTANCE BETWEEN TWO LINES

1st of all we shall find out shortest distance between two Parallel lines.

## Problem 1

Consider two parallel lines whose equations in vector form are given by

Now comparing these equations with standard form , and write

and vectors ,we get

Now applying this formula to find the shortest distance between two lines .
As it is clear from formula , we have to find cross product of and and then magnitude of vector =√(81)+(196)+(16)
=√293
Magnitude of = √49
= 7
Putting all these values in  the formula of Shortest Distance between two lines .
Now Find distance between two skew lines i.e. Lines which are not Parallel lines.

## Problem 2

Consider two parallel lines whose equations in vector form are given by
Now comparing these equations with standard form , and write

Now applying this formula to find the shortest distance between two lines

Now find cross product and then  magnitude of these two vectors

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## Solutions

Let ABCD be rectangle which is  inscribed in a given circle of radius ‘r’ And Let θ be the angle between side of rectangle and Diameter of given circle.

Therefore from right angled  Δ ABC ,

We have AB  = AC cosθ          ∵ AC = 2r
Let A(x) be the area of Rectangle ABCD
∴ A(x) = AB × BC
A = (2r cos θ)(2r sin θ )
A =  4r2 sin θ cos θ
A = 2r2  (2sin θ cos θ)
A = 2r2  (sin 2θ )

⇒ 2r2 2 (cos 2θ ) = 0 ,As r2 is constant
⇒cos 2θ = 0
⇒cos 2θ =cos (π/2)
⇒ θ = π/4

∴ A has Maximum value at θ = π/4

## Solution

Let us consider two numbers x and 16- x .
Then transforming our problem to mathematical form which says “sum of whose cube”  as follows
A (x) =   x3 + (16 - x)3…….. (1)
Differentiating both sides w .r. t  “x” , we get

X = 8
So  x  =  8 will be the 1st required numbers if Double derivatives of A  w. r. t  ‘x’ comes to be positive at x = 8.
Differentiate (2)  w. r. t. ‘x’  .

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## HOW TO SOLVE LINEAR EQUATIONS OF THREE VARIABLES BY MATRIX METHOD

How to find solutions of system of linear equations of three variables with  the help of Matrix Method. Let us solve three linear equations of three variables given below by Matrix Method. These are not simple linear equations to solve. But these equations can be made simple by following substitution .

After substitution , these equations are reduced to following simple linear equations
2x + 3y + 4z = 3
3x - 4y + 5z = 5
x + 2y - 3z = 6
To solve these equations by Matrix Method , Let us transform theses set of linear equations to Matrix form

AX = B,   and  =A-1 B---------------- (4)

where A is matrix comprises the coefficients  x , y and z respectively as shown in the matrix given below , And B is the matrix consist of constant terms on right hand side of these equations and X is the matrix of variables in the linear equations.

1st of all we have to find A- and then product of the matrices A-1 and  as mentioned in equation (4). But to find inverse of matrix A, its determinant value must be non zero   ( Note it) .
Let us find determinant value of Matrix A as follows :-

|A| = 2[(-4)(-3)–(5)(2)]-3[(3)(-3)-(5)(1)]+4[(3)(2)-(-4)(1)]

|A| = 2[12-10]-3[-9-5]+4[6+4]
|A| = 2 -3[-14]+4
|A| = 4 +42+40
|A| = 86
Since the value of Determinant of Matrix a is non zero.
⇒    A-1  exists .
To find inverse ,we have to find ad joint of Matrix A . And to find Ad joint of any Matrix , we have to find Co factor Matrix .

## What is Co Factor Matrix

Co factor Matrix can be obtained by the co factors of all the elements of Matrix written in same place  where the element was written originally.
Formula for Co factor of any element
Cij = (-1)i+jMij

### Co factor of A11

Co factor of A11   element can be calculated by eliminating 1st row and 1st column and solving the remaining determinant.
(-1)1+1M11
= 12 - 10 = 2

### Co factor of A12

Co factor of A12   element can be calculated   by eliminating 1st row and 2nd column and solving the remaining determinant.
(-1)1+2M12

= -1(-9 - 5 )= 14

### Co factor of A13

Co factor of A13   element  can be  calculated   by eliminating 1st row and 3rd column and solving the remaining determinant.
(-1)1+3M13
= 6 + 4 = 10

### Co factor of A21

Co factor of A21   element can be calculated   by eliminating 2nd row and 1st column and solving the remaining determinant.
(-1)2+1M11

= -1(- 9 - 8) = 17

### Co factor of A22

Co factor of A22   element  can be calculated   by eliminating 2nd row and 2nd column and solving the remaining determinant.
(-1)2+2M22
= - 6 - 4 = -10

### Co factor of A23

Co factor of A23   element can be calculated   by eliminating 2nd row and 3rd column and solving the remaining determinant.

(-1)2+3M23
= -1(4 - 3) = -1

### Co factor of A31

Co factor of A31   element can be calculated   by eliminating 3rd row and 1st column and solving the remaining determinant.

(-1)3+1M31
= 15 + 16 = 31

### Co factor of A32

Co factor of A32   element  can  be calculated   by eliminating 3rd row and 2nd column and solving the remaining determinant.

(-1)3+2M32
= -1(10 - 12) = 2

### Co factor of A33

Co factor of A33   element  can  be calculated   by eliminating 3rd row and 3rd column and solving the remaining determinant.

(-1)3+3M33
= - 8 - 9 = -17

## Co Factor Matrix of A

Now write all the co factors calculated in Matrix Form as follows

## Ad joint Matrix of A

To find the Ad Joint of  Co factor Matrix, transform 1st row to 1st column, 2nd row to 2nd column  and 3rd rows to 3rd column as follows

## Formula to Find   A-1

To Find the value of  A-1 ,Putting the value of Adj A in the formula given below.

## Find values of x ,y and z

Multiplying both the matrices which are on the right side.
Simplifying the matrix so obtained
Dividing each element by 86 to get matrix of 3×1 (i. e. Perform scalar multiplication of matrix ).

Using the equality of two matrices .We can write the values of x, y  and z respectively like this

x =277/86  , y = 4/86  and z = -77/86

Now the values of P, Q and R can be calculated by putting the values of x , y and z in equations (1) , (2 ) and (3) respectively.

P = 86/277  ,Q = 86/4   and R = -86/77

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## Verification of solution

Putting values of P , Q and R in given  equations , these values must satisfies  three given equations
From 1st equation

Similarly other two equations can be verified

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## Conclusion

I have discussed the method of solving the System of linear Equations with the help of Matrix Method , method to find inverse of a matrix ,How to find inverse of 3×3 Matrix,how to solve determinant,how to solve system of equation by matrix,matrix method of solving system of equations of three variables,If you liked the post Don't  forget to share it with your friends , And in case of any improvement please make use of Comment Box .

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