本人有數學不識“““幫助

1)3,6,12,24...........................
we have, first term a=3, common ratio R = 6/3 =2
the GENERAL TERM = aR^(n-1)= 3(2)^(n-1)
2)0.2,0.6,1.8,5.4.............
a= 0.2, R = 0.6/0.2 =3
the GENERAL TERM = aR^(n-1)= 0.2(3)^(n-1)

1)1+4+7+....TO30trems
first term a = 1, common different d= 4-1=3
THE SUM OF A.S.=(n/2)[2a +(n-1)d]
= (30/2)[2(1) +(30-1)(3)]=...=1335
2)80+76+72+......+40
a=80, d=76-80 = -4
let the number of terms be n,
40 = 80 +(n-1)(-4)
...
n=11
THE SUM OF THE FOLLOWING A.S.
=(n/2)[first term + last term]
= (11/2)[80+40]
=660


3.find the number of trems in the g.s.;1,3,9,....243
a=1, R=3/1 =3
(1)(3)^(n-1) = 243
(3)^(n-1) = 3^5
....
....
n= 6

4
.fid the gometric mean of -24 and -6
the gometric mean m = +/-√(-24)(-6)= +/-12

5.insert 2 geormetric means between -0.5 and 4
let the common ratio = R
then the four number -0.5, -0.5R, -0.5R^2, 4
the four term is -0.5R^3 = 4
R^3 = -8
R^3 = (-2)^3
R= -2
-0.5(-2)= 1 ,
-0.5(-2)^2 = -2
the 2 geormetric means between -0.5 and 4 are 1 and -2

6.1) find the sum of the following geomertic seric
1)1+(-2)+4+(-8).....to 15 terms
2)2+6+18+54+....+4374
1) the common ratio R = -2/1 =-2
the sum of the geomertic seric : a(1-R^n)/1-R
(1)[1-(-2)^15]/[1-(-2)]= .... = 10923

2) the common ratio R = 6/2 = 3
let the number of the term be n
2(3)^(n-1) = 4374
(3)^(n-1) = 2187
(3)^(n-1) = 3^7
n=8

the sum of the geomertic seric : a(R^n - 1)/R -1
(2)[3^8 - 1]/(3-1)=6560