2條數學問題

1. T(n) = a+(n+1)d
∵T(9) = 0
∴ a+(9-1)d = 0
i.e. a+8d = 0
For the 29th term,
T(29) = a+(29-1)d
= a+28d
= a+8d+20d
= 20d
For the 19th term,
T(19) = a+(19-1)d
= a+18d
= a+8d+10d
= 10d
∴T(29) = 2xT(19)

2.S(n) = a(1-r^n)/(1-r)
∵S(5) = 121
∴a(1-r^5)/(1-r) =121 -(1)
∵S(10) - S(5) = 29403
∴a(1-r^10)/(1-r) - 121 = 29403
a(1-r^10)/(1-r) = 29524 -(2)

(2)/(1) : (1-r^10)/(1-r^5) = 244
1-r^10 = 244 - 244(r^5)
r^10 - 244(r^5) + 243 = 0
[(r^5)-1][(r^5)-243] = 0
r^5 = 1 or r^5 = 243
r = 1 (rejected,∵when r = 1,a(1-r^n)/(1-r) will be undefined) or r = 3
∴ r = 3
sub r = 3 into (1),
a(1-3^5)/(1-3) = 121
a(1-243) = -242
a = 1
∴ The first term = 1, common ratio = 3

1.
等差數列第n項的值為 a + (n - 1)d

when n = 9,
a + 8d = 0

when n = 19,
the value is a + 18d = (a + 8d) + 10d = 10d

when n = 29,
the value is a + 28d = a + 8d + 20d = 20d = 2 x 10d

so 數列的第29項為第19項的兩倍


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2.
等比數列的首n項之和為
[a(r^n - 1)] / (r - 1)

when n = 5,
[a(r^5 - 1)] / (r - 1) = 121 (Equation 1)


when n = 10,
[a(r^10 - 1)] / (r - 1) = 121 + 29403 = 29524 (Equation 2)


Equation 2 / Equation , we have

(r^10 - 1) / (r^5 - 1) = 244

r^10 - 1= 244r^5 - 244

r^10 - 244r^5 + 243 = 0

(r^5)^2 - 244r^5 + 243 = 0

r^5 = 243 or 1 (rejected)

r = 3

put r = 3 into [a(r^5 - 1)] / (r - 1) = 121 (Equation 1)

[a(3^5 - 1)] / (3 - 1) = 121

a = 1

so,
數列的首項是1, 公比是3.