Question

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

Answer 1

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 $n={\displaystyle \frac{l-a}{d}}+1$ .

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
$\Rightarrow -43-(-44)=-43+44=1$
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
.$\Rightarrow n={\displaystyle \frac{l-a}{d}}+1$.
Where a is first term, d is common difference,l is last term of an A.P
Substituting the values we get
$$\begin{gathered} \Rightarrow n={\displaystyle \frac{45-(-44)}{1}}+1 \\ \Rightarrow n={\displaystyle \frac{45+44}{1}}+1 \\ \Rightarrow n={\displaystyle \frac{89}{1}}+1 \\ \Rightarrow n=89+1 \\ \Rightarrow n=90 \end{gathered}$$
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 $a_n=a+(n-1)d$
$$\begin{gathered} \Rightarrow 45=-44+(n-1)(1) \\ \Rightarrow 45+44=(n-1) \\ \Rightarrow 89+1=n \\ \Rightarrow 90=n \end{gathered}$$