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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,.....,495In 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=a2−a1=319−308=11We know that the general term of an arithmetic progression with first term a and common difference d is Tn=a+(n−1)d, therefore, with a=308,d=11 and Tn=495, we haveTn=a+(n−1)d⇒495=308+(n−1)11⇒495=308+11n−11⇒495=297+11n⇒11n=495−297⇒11n=198⇒n=19811⇒n=18We also know that the sum of an arithmetic series with first term a and common difference d is Sn=n2[2a+(n−1)d]Now to find the sum of series, substitute n=18,a=308 and d=11 in Sn=n2[2a+(n−1)d] as follows:S18=182[(2×308)+(18−1)11]=9[616+(17×11)]=9(616+187)=9×803=7227Hence, the sum is 7227.

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