Question

a) What is the Remainder obtained when the terms of the Arithmetic sequence 301, 308, 315.... is divided by 7. b) Find the sum of natural numbers between 300 and 500 which leaves a reminder of 2 on division by 7. ​

Answer 1

Divisors of 7 between 100 and 500 are,

105,112,119,126,…..,497.

This is in A.P.

Here a=105, d=7 and Tn=l=497.

Tn=a+(n-1)d

==>497=105+(n-1)*7

==>(n-1)*7=392.

==>n=56+1

n=57.

Sn=[a+l]*(n/2)

Sn=602*57 /2

Sn=17,157.

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