Curly213 Posted February 9, 2009 Share Posted February 9, 2009 Could someone help me with this sequences and series ex. 1. How many numbers between 200 and 3600 inclusive are divisible by 4, 5 and 6? 57 2. Find the sum of all two digit numbers which leave a remainder of 3 when divided by 7? 676 3. In a series of 10 consecutive integers what is the sum of the first 5 integers if the sum of the last 5 is 375? 350 4. If the sum of 5 consecutive integers is 95 what is the sum of the first and last integer? 38 5. What is the sum of the first 50 positive numbers that when divided by 8 leave a remainder of 4? 10000 6. If the sum of 10 consecutive integers is 1005 how many of the numbers are prime? 3 7. What is the sum of all integers between 273 and 297 inclusive? Quote Link to comment Share on other sites More sharing options...
tpcool Posted February 9, 2009 Share Posted February 9, 2009 Thanks this is a good chunk of questions on sequence and series The document at this URL has good info on APs and GPs. It should help us in solving all such kind of questions http://www.mathcentre.ac.uk/resources/leaflets/mathcentre/business/arith_and_geom_progressions.pdf 1. How many numbers between 200 and 3600 inclusive are divisible by 4, 5 and 6? 57 Take lcm of 4,5,6 -> 60 First number divisible by 60 = 240 Last number diviisble by 60 = 3600 This forms an arithmetic progression with common difference 60 nth term formula = a + (n-1)d 3600 = 240 + (n-1)60 n = 57 2. Find the sum of all two digit numbers which leave a remainder of 3 when divided by 7? a = 10 l = 94 d = 7 no. of terms. use nth term formula on last term 94 = 10 + (n-1)*7 n = 17 average = (a + l)/2 = (10 + 94)/2 = 52 sum = average * no. of terms = 52 * 17 = 884 [we could also direct formula for sum = n * (2a + (n-1)d)/2 ] 3. In a series of 10 consecutive integers what is the sum of the first 5 integers if the sum of the last 5 is 375? Find out the sixth term (which is the first term for the last 5) 375 = 5/2(2a + 4) -> a = 73 -> first term for the serie is 68 sum of first five = 5/2 * (2 * 68 + 4) = 350 4. If the sum of 5 consecutive integers is 95 what is the sum of the first and last integer? 95 = 5a + 10 -> a = 17 -> l = 17 + 4 a+l = 38 5. What is the sum of the first 50 positive numbers that when divided by 8 leave a remainder of 4? sum = 50/2 (24 + 49*8) = 10400 6. If the sum of 10 consecutive integers is 1005 how many of the numbers are prime? first term = 96 last term = 105 list out numbers. primes = 97,101,103 7. What is the sum of all integers between 273 and 297 inclusive? sum = (273 + 297)/2 * 25 = 7125 Quote Link to comment Share on other sites More sharing options...
ashk29 Posted March 15, 2009 Share Posted March 15, 2009 if the number must be divisible by 4,5and 6 then it should be the LCM of 4,5and 6 that is 60 so the number of multiples of 60 between 200 and 3600 No.= 360-24/6 +1 = 56+1 =57 Quote Link to comment Share on other sites More sharing options...
ashk29 Posted March 15, 2009 Share Posted March 15, 2009 2.0 Find the sum of all two digit numbers which leave a remainder of 3 when divided by 7 so the number is in the form of 7n+3 , n an integer so min n = 1 and ma n =13 so sum them up ans= 7*13*14/2 + 3*13 = 13*52 = 676 Quote Link to comment Share on other sites More sharing options...
ashk29 Posted March 15, 2009 Share Posted March 15, 2009 In a series of 10 consecutive integers what is the sum of the first 5 integers if the sum of the last 5 is 375 Consecutive integers --OK sum of the last 5 = 375 then sum of the first 5 = 350 No need to find a Its very very simple if i write the number series I get a, a+1,a+2,a+3...........................a+9 so sum is 10a+sum 1to 9 we break the series in to to 5a+sum 1to 4 + 5a+sum 5to 9 o whats the difference , the difference is 25 so deduct 25 from 375 you get 350 no need for fancy equations just logic that's it:) Quote Link to comment Share on other sites More sharing options...
ashk29 Posted March 15, 2009 Share Posted March 15, 2009 If the sum of 5 consecutive integers is 95 what is the sum of the first and last integer? Important words are consecutive integers Using our common sense we see that 95 is divisible by 19 that is 95=19*5 that means if we take 19 as the base point we have (or as the median) the 5 means the numbers(5 consecutive integers) NOTE : 19*5 = 19+19+19+19+19 OK :grad: I have written it as a sum of 5 integers so what can I do ; I can do this 19-2,19-1,19,19+1,19+2 when we add the + and - get canceled off and there are only 5 integers so the first must be 17 and the last must be 21 then the answer is 17+21 = 38 NO equation solving needed !!!!!!!!!!!! Quote Link to comment Share on other sites More sharing options...
ashk29 Posted March 15, 2009 Share Posted March 15, 2009 What is the sum of the first 50 positive numbers that when divided by 8 leave a remainder of 4 The first number is 4 the series is 4 ,12,20,28,............, we know the first and the last number and we know that there are 50 numbers so the answer is 10000 Quote Link to comment Share on other sites More sharing options...
ashk29 Posted March 15, 2009 Share Posted March 15, 2009 If the sum of 10 consecutive integers is 1005 how many of the numbers are prime? Use the same logic as the previous sum ( take 100 as base) primes 91,101,103 Quote Link to comment Share on other sites More sharing options...
kushagra452 Posted March 16, 2009 Share Posted March 16, 2009 Q5) a= 4 l=(8*49+4) (since we need to consider the first 50 positive numbers) d= 4 therefore n= 99 using: Sn = n/2 * {2a+(n-1)d} we get 200*99= 19800 Is this correct?? Quote Link to comment Share on other sites More sharing options...
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