the sum of 10 consecutive integers starting with 11 equals the sum of 5 consecutive integers starting with
A)22
B)29
C)31
D)33
Math question ?
2020-08-07 12:30 pm
回答 (2)
2020-08-07 12:39 pm
Let a be the integer that the 5 consecutive starting with.
Sum of n integers starting with a = [2a + (n - 1)] n / 2
[2a + (5 - 1)] × 5 / 2 = [2×11 + (10 - 1)] × 10 / 2
2a + 4 = (22 + 9) × 2
2a + 4 = 62
2a = 58
a = 29
The answer: B) 29
Sum of n integers starting with a = [2a + (n - 1)] n / 2
[2a + (5 - 1)] × 5 / 2 = [2×11 + (10 - 1)] × 10 / 2
2a + 4 = (22 + 9) × 2
2a + 4 = 62
2a = 58
a = 29
The answer: B) 29
2020-08-07 12:35 pm
S = 11 + 12 + 13 + ... + 20
t[1] = 11
t[10] = 20
S = (10/2) * (11 + 20)
S = 5 * 31
S = 155
T = n + n + 1 + n + 2 + n + 3 + n + 4
T = (5/2) * (n + n + 4)
T = (5/2) * (2n + 4)
T = 5 * (n + 2)
T = 5n + 10
5n + 10 = 155
n + 2 = 31
n = 29
t[1] = 11
t[10] = 20
S = (10/2) * (11 + 20)
S = 5 * 31
S = 155
T = n + n + 1 + n + 2 + n + 3 + n + 4
T = (5/2) * (n + n + 4)
T = (5/2) * (2n + 4)
T = 5 * (n + 2)
T = 5n + 10
5n + 10 = 155
n + 2 = 31
n = 29
收錄日期: 2021-04-18 18:35:39
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