1st problem :::

If 1+2=5 and 3+4=12,then 10+13 = ?

2nd problem :::

If 5=5 , 10=55 , 15=610 then 20= ?

if 5=10 , 10=55 ,15=610 than 20=?

if 5=10 , 10=55 ,15=610 than 20=?

if 5=10 , 10=55 ,15=610 than 20=?

if 5=10 , 10=55 ,15=610 than 20=?

Guest Jun 30, 2014

#6**+5 **

You just have to findout 1st 13 prime numbers.

1st prime number is 2 and 2nd is 3 so 1 + 2 = 5 (2 + 3).

3rd is 5 and 4th is 7 then 3 + 4 = 12 ( 5 + 7).

And next same like this.

Prime series : [ 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61 ]

so answer is 10 + 13 = (29+41)=70

Guest Jul 1, 2014

#1**+5 **

The second one is the Fibonacci Series......0,1,1,2,3,5,8,13,21,34,55,89,144.......

The nth term ...starting with the second term (0 is the 0th term) is just the sum of the previous two

The nth term ( Fibonacci Number) is given by

F(n) = [ (1 + √5 )^{n} - (1- √5 )^{n} ] / (2^{n}√5)

F(20) = 6765

CPhill
Jun 30, 2014

#3**+5 **

Let's assume that we have an arithmetic series here (I don't know if we do, or not)

An arithmetic series is given by

a_{n+1} = a_{n} + (n-1)d

Where a_{n+1} is the term following a_{1}.....n represents the nth term of the series, and d is the common difference between successive terms

Let's call the first term a_{1}. And the second term(a_{2}) is just a_{1} + (2-1)d = a_{1} + d

So the sum of the first two terms is just

a_{1}+ (a_{1} + d) = 5 or 2a_{1} + d = 5 .....call this Equation 1

And the third term (a_{3}) is just .... a_{1} + (3-1)d = a_{1} + 2d

And the fourth term(a_{4}) is just..... a_{1} + (4-1)d = a_{1} + 3d

And the sum of these two terms = 12...... so

a_{1} + 2d + a_{1} + 3d = 12 or 2a_{1} + 5d = 12 ...... call this Equation 2

Now multiplying Equation 1 through by -1 on both sides and adding it to Equation 2, we get

4d = 7 so d = 7/4

And substituting for d in Equation 1, we have

2a_{1} + 7/4 = 5

2a_{1} = 13/4

a_{1} = 13/8 and that's the first term.....Now, lets check our results

a_{2} = a_{1} + (2-1)(7/4) = 13/8 +7/4 = 27/8

So a_{1} + a_{2}= 13/8 + 27/8 = 40/8 = 5 .....and that's OK

And the sum of a_{3} and a_{4} is given by Equation 2 = 2a_{1} + 5d = 12

So we have

2(13/8) + 5(7/4) =

26/8 + 35/4 =

26/8 + 70/8 =

96/8 = 12 .....and that's OK, too

So our series (so far) is just

13/8, 27/8, 41/8, 55/8

And the sum of the 10th and 13th terms is

[a_{1} + (9)d] + [a_{1} + (12)d] = 2a_{1} + 21d = 2(13/8) + 21(7/4) = 26/8 + 294/8 = 320/8 = 40

CPhill
Jun 30, 2014

#5**0 **

You don't know the half of it, zegroes....I spent all afternoon helping a girl with an Algebra 2 Test and a Trig Test...........

In short....I'm fried !!!!

CPhill
Jul 1, 2014

#6**+5 **

Best Answer

You just have to findout 1st 13 prime numbers.

1st prime number is 2 and 2nd is 3 so 1 + 2 = 5 (2 + 3).

3rd is 5 and 4th is 7 then 3 + 4 = 12 ( 5 + 7).

And next same like this.

Prime series : [ 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61 ]

so answer is 10 + 13 = (29+41)=70

Guest Jul 1, 2014