Maths Question 非常急

1
S(8) = n/2 [2a + (n-1) d]
336 = 8/2 [2a + 7(-4)]
336 = 4 (2a - 28)
84 = 2a - 28
112 = 2a
a = 56
First term = 56

2
(a) First term = T[1] = 55 - 2(1) = 53
T[2] = 55 - 2(2) = 51
Common difference = T[2] - T[1] = 51 - 53 = -2
(b) Put T[n] < 0
55 - 2n < 0
55 < 2n
27.5 < n
Therefore, n = 28
First negative term= T[28] = 55 - 2(28) = -1
(c) S[n] = n/2 [2a + (n-1)d]
= n/2 [2(53) + (n-1)(-2)]
= n/2 [106 - 2n + 2]
= n/2 [108 - 2n]
= - (n^2 - 54n)
= - [n^2 - 54n + (27)^2] + (27)^2
= - (n - 27)^2 + 729
Therefore S(n) is maximum when n = 27
Maximum value = 729

3
First term a = 2
Common ratio r = -4
T(n) = ar^(n-1)
-2048 = 2(-4)^(n-1)
-1024 = (-4)^(n -1)
n-1 = 5
n = 6
Number of terms = 6
S(6) = a[r^n - 1]/(r - 1)
= 2 [ (-4)^6 - 1]/5
= 2(4095)/5
= 1638
Therefore, sum of terms is 1638

1.d=-4
S(n)=336
S(n)=8/2 { 2a+(8-1)(-4) }=336
2a-28=84
a=56

2.T[n]=55-2n
a.the first term =T(1)=55-2=53
b.common difference=T(2)-T(1)=51-53=-2

c.T[n]=55-2n&gt;0
n&lt;27.5
n=27
the maximum value of T[n]=55-2(27)=1
the maximum value of S[n]
=27/2[53+1]
=729

3.a=2,r=-4
T(n)=ar^n=-2048
2(-4)^n=-2048
(-4)^n=(-4)^5
n=5
sum of the terms={ 2 *[ 1-(-4)^5] } / (1+4)
=410