What is the next number in series 4 8 15 25 38?

Answer 1

n=number of term (ie 4 has n=1, 8 has n=2)

t=value or term number 'n' (n=3 has t=15)

Easy way:

4 --> 8 =+4

8 --> 15 = +7

15 --> 25 =+10

25 --> 38 = +13

Note: increases are also increasing by 3

Note: next increase should be +16

38+16=54

4 --> 8 = +4

4 --> 15 =+11

4 --> 25 =+21

4 --> 38 =+34

+4 = 4+0

+11= (4+0)[previous increase]+(4+0+3) [initial increase plus 3]

+21= ((4+0)+(4+0+3))[previous increase]+((4+0+3)+3)[initial increase plus 3 again]

+34= ((4+0)+(4+0+3))+((4+0+3)+3)[previous increase]+4+0+3+3+3

+4= 4*1+3*0

+11 = 4*2+3*1

+21 = 4*3+3*3

+34 = 4*4+3*6

Note: 0,1,3,6 are the sums of an arithmetic series that increases by 1 and starts at 0 (ie. 0=0, 1=0+1, 3=0+1+2)

S=(n/2)(2n(initial)+(n-1)d)

where n = term number

where n(initial) = n where t=1

where d = increase in series

S=(n/2)(2*0+(n-1)*1)

S=(n/2)(n-1)

Therefore additions:

4n+3*(n/2)(n-1)

So, for the first increase (ie 4 --> 8). The increase = 4*1+3*0.5*0=4

1st increase = 4*1+3*0.5*0=4

2nd increase = 4*2+3*1*1=11

3rd increase = 4*3+3*1.5*2= 21

Note if you add 4(the first term of the initial series you get the right answers but shifted one term:

ie:

with:

4+4n+3*(n/2)(n-1) = 4(n+1)+3*(n/2)(n-1)

t1=8

t2=15

t3=25

t4=38

Merely shift the n values to correct (ie. n --> n-1, n-1 --> n-2 etc.):

t = 4n+3((n-1)/2)(n-2)

t6= 4*6+3*2.5*4=24+30=54