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The least number divisible by both 6 and 8 is 24. Since the question is asked between 100 and 1000, the least number which is both a multiple of 8 and 6 above 100 is 120. The sequence ends in 984. Thus, here, a = 120 ; last term, l = 984, and the sequence's terms increase by 24, i.e. d=24. Thus, n = (l -a)/d +1 = (984-120)/24 +1 = 37. Sum of all these terms = n/2 [a +l] = 37/2 [ 1008 ] = 20424.

LCM of 6 and 8 is 24 $$ 1000/24 = 41.66 => 41x24 = 984 $$ 100/24 = 4.16 => 5x24=120 and 4x24=96 $$ $$ Now, $$ S = 120 + 144 + ... ... ... + 984 $$ = 24 x (5 + 6 + 7 + ... ... ... + 41) $$ = 24 x [(1+2+3+4+... ... ... + 41) − (1+2+3+4)] $$ = 24 x [(½).41.42 − (½).4.5] $$ = 24 x (861 − 10) $$ = 20424

The least common multiple of 6 and 8 is 24. Therefore, we know that all of the terms in our series are multiples of 24. The first multiple within the range is 120; the last is 984. We can set up a series in sigma notation as such $sum_{i=1}^{37} (24i+96)$ . Using the formulas for sums (here, k[n][n+1]\2 and k[n]) we know that this is equal to (24)(37)(38)/2+96(37) = 20424