find the sum 46+47+48+49...... +114+115
Answers
Given a series (45+46+47+………+113+114+115)
We need to find the sum of the given series
Solution
The given series is in arithmetic progression
So the sum of the series in arithmetic progression is given by
Sum=n/2(a+an) ——————-(i)
Where
n is number of terms
a is the first term
an is the last term of the series
We need to find the value of n, the number of terms first
We know that
an=a+(n-1)d ——————–(ii)
From the series an = 115, a= 45
Substituting the value of an = 115 and a=45 in equation (ii) we get,
115=45+(n-1)1
115=45+n-1
115=44+n
n=115-44
n=71
Substituting the value n=711 in equation (i) we get,
Sum=n/2(a+an)
=71/2(45+115)
=71/2×160
=71×80
=5680
Therefore the sum of the series (45+46+47+………+113+114+115) = 5680
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