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

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 = \dfrac{{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
$
   \Rightarrow n = \dfrac{{45 - ( - 44)}}{1} + 1 \\
   \Rightarrow n = \dfrac{{45 + 44}}{1} + 1 \\
   \Rightarrow n = \dfrac{{89}}{1} + 1 \\
   \Rightarrow n = 89 + 1 \\
   \Rightarrow n = 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 ${a_n} = a + (n - 1)d$
$
   \Rightarrow 45 = - 44 + (n - 1)(1) \\
   \Rightarrow 45 + 44 = (n - 1) \\
   \Rightarrow 89 + 1 = n \\
   \Rightarrow 90 = n \\
$