Question

The sum of multiples of 7 lying between 101 to 999.
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Answer 1

Step-by-step explanation:

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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Answer 2

Answer:

105,112,119,126…. 994

a=105

an=994

d=7

an=a+(n-1)d

994=105+(n-1)7

994-105=7n-7

889+7=7n

896/7=n

n=128

an=n/2(105+994)

sn =128/2(1099)

sn=64×1099

sn=70336