Question

Find the sum of the sequence.43 +44 +45 + 46+ ... + 116

Answer 1

The sequence is an arithmetic progression. The common difference is 1 while the first term is 43 .

d = common difference

a = first term

Since we were given last term we use the following formula

$\begin{gathered} sum=\text{ }\frac{n}{2}(a+l) \\ \end{gathered}$

But we need to find number of term which is n

$\begin{gathered} nth\text{ term = a+(n-1)d} \\ a\text{ = 43} \\ d\text{ = 1} \\ nth\text{ term = 116} \\ 116\text{ = }43+(n-1)1 \\ 116\text{ = 43+n-1} \\ 116-42\text{ = n} \\ n\text{ = 74} \end{gathered}$$\begin{gathered} sum=\text{ }\frac{n}{2}(a+l) \\ sum=\text{ }\frac{74}{2}(43+116) \\ sum=\text{ }37(159) \\ sum=\text{ }5883 \end{gathered}$