Find the sum of first 40 positive integers divisible by 6. Also find the sum of first 20 positive integers divisible by 5 or 7.
Answer (1)
10th Mar, 2020
1) Positive integers divisible by 6 are 6, 12, 18, 24,…. Since difference is same, it is an AP We need to find sum of first 40 integers We can use formula Sn = /2 (2a + (n – 1) d) Here , n = 40 , a = 6 & d = 12 – 6 = 6 Putting values in formula Sn = /2 (2a + (n – 1) d) Sn = 40/2(2×6+(40−1)×6) Sn = 20 (12 + 39 × 6) Sn = 20 (12 + 234) Sn = 20 × 246 Sn = 4920 Therefore, the sum of first 40 multiples of 6 is 4920
2)
Assuming that we are only adding whole numbers that are evenly divisible by 5 or 7, (or both) we consider the following multiples of 5 and 7:
- 5,10,15,20,25,30, 35 ,40,45,50,55,60, The sum of the preceding integers may be written as : 5*(1+2+3+4+5+6+7+8+9+10+11+12) = 5*78= 390
- 7,14,21,28, 35, 42,49,56,63, Again, the sum of the preceding values can be written similarly as: 7*(1+2+3+4+5+6+7+8+9) = 7*45= 315. Of course, since we’ve included the integer 35 twice (it’s the LCM of 5 and 7), it must be subtracted from the 2nd sum, so that 315 - 35 = 280
- Now we have the 1st 12 multiples of 5 (including 35) and the 1st 8 multiples of 7 (excluding 35) with no other common multiples. Adding 280 and 390 gives us 670 as the total sum.
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