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What is the greatest number of consecutive integers whose sum is 45?

Answer
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Hint: When we add the numbers from -44 to 44 we get the sum to be 0 and adding the next number 45 to it we get the sum to be 45. Hence the numbers -44 , -43 , -42 , …… , 42 , 43 , 44 , 45 form an arithmetic sequence with the common difference 1 . We need to find the number of terms in the sequence which can be obtained by using the formula n=lad+1 .

Complete step-by-step answer:
Step 1:
We are asked to find the greatest number of consecutive integers which add up to 45
Adding the numbers from -44 to 45 we get 45
That is when we add numbers -44 , -43 , -42 , …… , 42 , 43 , 44
We get zero and adding the next number 45 we get the sum to be 45
Step 2
Now we can see that -44 , -43 , -42 , …… , 42 , 43 , 44 , 45 forms an arithmetic sequence
The common difference can be obtained by subtracting the second term from the first term
43(44)=43+44=1
Now we need to find the number of terms in this sequence
We have a = - 44 , d = 1 , l = 45
So the number of terms is given by
.n=lad+1.
Where a is first term, d is common difference,l is last term of an A.P
Substituting the values we get
n=45(44)1+1n=45+441+1n=891+1n=89+1n=90
Therefore the greatest number of consecutive terms that sum to 45 is 90.

Note: To find the value of n we can also use the formula an=a+(n1)d
45=44+(n1)(1)45+44=(n1)89+1=n90=n


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