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Question

Find the sum of all natural numbers between 300 and 500 which are divisible by 11

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Solution

The natural numbers between 300 and 500 which are divisible by 11 are as follows:

308,319,330,.....,495

In the above sequence, the first term is a1=308, second term is a2=319 and the nth term is Tn=495.

We find the common difference d by subtracting the first term from the second term as shown below:

d=a2a1=319308=11

We know that the general term of an arithmetic progression with first term a and common difference d is Tn=a+(n1)d, therefore, with a=308,d=11 and Tn=495, we have

Tn=a+(n1)d495=308+(n1)11495=308+11n11495=297+11n11n=49529711n=198n=19811n=18

We also know that the sum of an arithmetic series with first term a and common difference d is Sn=n2[2a+(n1)d]

Now to find the sum of series, substitute n=18,a=308 and d=11 in Sn=n2[2a+(n1)d] as follows:

S18=182[(2×308)+(181)11]=9[616+(17×11)]=9(616+187)=9×803=7227
Hence, the sum is 7227.

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