Find the sum of first 40 positive integers which give a remainder 1 when divided by 6
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the first positive integer which divided by 6 gives the remainder 1 is 7 .
likewise the 40th terms will be : a+ (n-1) d
7 + (40-1)6
= 7+234
=241
THEREFORE sum of 40 terms is
N/2(a+l)
= 40÷2 (241 +7)
= 20 × 248
=4960
therefore the sum of first 40 terms which give the remainder 1 when divided by 6 is 4960
likewise the 40th terms will be : a+ (n-1) d
7 + (40-1)6
= 7+234
=241
THEREFORE sum of 40 terms is
N/2(a+l)
= 40÷2 (241 +7)
= 20 × 248
=4960
therefore the sum of first 40 terms which give the remainder 1 when divided by 6 is 4960
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