超簡單的數學題4! 教教我

1)
a = 80
d = -7
Sum = -1881
Thus,
-1881 = [ a + a + (n-1)d ] * n/2
-3762 = n[ 2(80) + (n-1)(-7) ]
-3762 = 160n + (-7)(n^2-n)
-3762 = 160n - 7n^2 + 7n
-7n^2 + 167n + 3762 = 0
7n^2 - 167n - 3762 = 0
(7n+99)(n-38) = 0
n = -99/7 (rej) or n=38
Number of terms is 38

2)
a = 40
d = -3
Sum = -740
Thus,
-740 = [ a + a + (n-1)d ] * n/2
-1480 = n[ 2(40) + (n-1)(-3) ]
-1480 = 80n - 3n^2 + 3n
-1480 = -3n^2 + 83n
3n^2 - 83n - 1480 = 0
(3n+37)(n-40) = 0
n = -37/3 (rej) or n=40
Number of terms is 40

3a)
The 5th term = a + (5-1)d = a+4d
The 10th term = a + (10-1)d = a+9d
5th term = 28
-> a+4d = 28 -- (1)
10th term = 53
-> a+9d = 53 -- (2)
(2) - (1)
a+9d - (a+4d) = 53-28
a + 9d - a - 4d = 25
5d = 25
d = 5
Put d=5 into (1)
a + 4(5) = 28
a + 20 = 28
a = 8
Thus,
a=8 and d=5

3b)
The sum of first 15 terms
= [ a + a + (15-1)d ] * 15/2
= (2a + 14d)*15/2
= [ 2(8) + 14(5) ] * 15/2
= (16+70)*15/2
= 86*15/2
= 645

4a)
a = 123
d = -13
a + (n-1)d > 0
123 + (n-1)(-13) > 0
123 - 13n + 13 > 0
136 - 13n > 0
136 > 13n
n < 10.46
Thus, n must be 10
The smallest positive term
= 123 + (10-1)(-13)
= 123 + 9(-13)
= 123 - 117
= 6

4b)
The sum of 10 terms
= [ 123 + 6 ] * 10/2
= 129 * 5
= 645

5)
T(n) = 3n - 11
1st term = 3(1)-11 = -8
2nd term = 3(2) - 11 = -5
3rd term = 3(3) - 11 = -2
.
.
.
a = -8
d = 3
The sum = 1505
Thus,
1505 = [ a + a + (n-1)d ] * n/2
3010 = [ 2*(-8) + 3n - 3 ] * n
3010 = -16n + 3n^2 - 3n
3010 = -19n + 3n^2
3n^2 - 19n - 3010 = 0
(3n+86)(n-35) = 0
n = -86/3 (rej) or n=35
Number of terms = 35