find sum of multiples of 7 lying between 103 and 999
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The AP is 105,112,119,126...........987,994. Since the first term is a and the last is a+(n-1)d, their difference divided by d gives us (n-1). So 994-105 = 889 when divided by 7 gives us 127. If (n-1) is 127, n is 128. Sum of terms is (n/2)(a+l) where l is the last term. The answer is (128/2)(105 + 994) = 70336.
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