Question

Find the sum of the given series \[\left( {45 + 46 + 47 + ...... + 113 + 114 + 115} \right)\]

Answer 1

Hint: A sequence of numbers is called an Arithmetic progression if the difference between any two consecutive terms is always the same. In simple terms, it means that the next number in the series is calculated by adding a fixed number to the previous number in the series.

Complete step-by-step answer:
Fact about Arithmetic Progression :
Initial term: In an arithmetic progression, the first number in the series is called the initial term.
Common difference: The value by which consecutive terms increase or decrease is called the common difference.
The behavior of the arithmetic progression depends on the common difference d. If the common difference is positive, then the members (terms) will grow towards positive infinity or negative, then the members (terms) will grow towards negative infinity.

Given the series \[\left( {45 + 46 + 47 + ...... + 113 + 114 + 115} \right)\]
We need to find the sum of the given series
The given series is in arithmetic progression
So the sum of the series in arithmetic progression is given by
Sum \[ = \dfrac{n}{2}({a_1} + {a_n})\]
Where \[n\] is number of terms
\[{a_1}\] is the first term
\[{a_n}\] is the last term of the series
We need to find the value of \[n\] , the number of terms first
We know that
\[{a_n} = {a_1} + (n - 1)d\]
From the series we have \[{a_n} = 115,{a_1} = 45\] and \[d = 1\]
Substituting these values we get ,
\[115 = 45 + (n - 1)(1)\]
Which on simplification gives us ,
\[n = 115 - 45 + 1\]
Which gives us,
\[n = 71\]
Now Substituting the value of \[n\] i.e. \[n = 71\] in the equation of sum of the series we get ,
\[ = \dfrac{{71}}{2}(45 + 115)\]

\[ = \dfrac{{71}}{2}(160)\]
Which on simplification gives us ,
\[ = 71 \times 80 = 5680\]
Therefore the sum of the given series is \[5680\].
So, the correct answer is “\[5680\]”.

Note: A sequence of numbers is called an Arithmetic progression if the difference between any two consecutive terms is always the same. The behavior of the arithmetic progression depends on the common difference d. If the common difference is positive, then the members (terms) will grow towards positive infinity or negative, then the members (terms) will grow towards negative infinity.