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# number Series problem ..... need answer >asap<

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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=?
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

Jul 1, 2014

#1
+94558
+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 ] / (2n√5)

F(20) = 6765

Jun 30, 2014
#2
0

Dear cphill,

Jun 30, 2014
#3
+94558
+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

an+1 = an + (n-1)d

Where an+1 is the term following a1.....n represents the nth term of the series, and d is the common difference between successive terms

Let's call the first term a1. And the second term(a2) is just a1 + (2-1)d = a1 + d

So the sum of the first two terms is just

a1+ (a1 + d) = 5   or     2a1 + d = 5    .....call this Equation 1

And the third term (a3) is just ....  a1 + (3-1)d =   a1 + 2d

And the fourth term(a4) is just.....    a1 + (4-1)d  = a1 + 3d

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

a1 + 2d + a1 + 3d  = 12    or    2a1 + 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

2a1 + 7/4 = 5

2a1 = 13/4

a1 = 13/8   and that's the first term.....Now, lets check our results

a2 = a1 + (2-1)(7/4)  = 13/8 +7/4  =  27/8

So    a1 + a2= 13/8 + 27/8 = 40/8 = 5  .....and that's OK

And the sum of a3 and a4  is given by Equation 2 = 2a1 + 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

[a1 + (9)d] + [a1 + (12)d] = 2a1 + 21d =  2(13/8) + 21(7/4) =  26/8 + 294/8 = 320/8 = 40

Jun 30, 2014
#4
+4151
0

Hey chris dosent your fingers get tired!!!!???Lol

Jun 30, 2014
#5
+94558
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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 !!!!

Jul 1, 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
#7
+94558
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Ah....your answer was certainly "easier," Anonymous!!!.......but mine was  pretty interesting, too........if I do say so myself !!!

Jul 1, 2014