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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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