Question

Following is the development of a formula for the sum of n consecutive integers use the formula to find the sum of the numbers 101-110

Answer 1

The sum of the numbers 101-110 can be calculated using the formula for the sum of n consecutive integers. Substituting the values into the formula S = n/2 × (first term + last term) results in a sum of 1055.

Step-by-step explanation:

The sum of n consecutive integers can be found using the formula S = n/2 × (first term + last term). In the case of the numbers 101-110, there are 10 consecutive numbers (n = 10). The first term (a1) is 101 and the last term (an) is 110. Substituting these values into the formula, we get:

S = 10/2 × (101 + 110) = 5 × 211 = 1055.

Therefore, the sum of the numbers from 101 to 110 is 1055.

Answer 2

Note that the formula n(n+1)/2 is used only for adding n consecutive integers starting from 1.
A problem may not directly ask you for this but if you can break it down such that you have to find the sum of 'n consecutive integers starting from 1' then you can use this formula.

In the even integers questions, you may be required to find the sum of first 10 even integers.
2 + 4 + 6 + ... + 18 + 20
Take 2 common, 2*(1 + 2 + 3 + ...10)
To find the sum of the highlighted part, we can use the formula. Then we can multiply it by 2 to get the required sum.

Note that for odd integers, you cannot directly use this formula.
Sum the first 10 odd integers
1 + 3 + 5 + 7+...+19
But you can still make some modifications to find the sum.

1 + 3 + 5 + 7+...+19 = (1 +2+ 3 + 4+5 + 6+ 7+...+19 + 20) - (2 + 4+ 6+...20)
We know how to sum consecutive integers.
(1 +2+ 3 + 4+5 + 6+ 7+...+19 + 20) = 20*21/2
(2 + 4+ 6+...20) = 2 * (10*11)/2 = 10*11 (as before)

So 1 + 3 + 5 + 7+...+19 = (20*21/2) - (10*11) = 100

The direct formula of sum of n consecutive odd integers starting from 1 = n^2
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