46
2 + 4 + 6 + ... + 44 + 46 + 48 = 600
2304Sum of an arithmetic progression = n/2 (2a + (n-1)d) where n is the number of items, a is the first item and d is the difference between them:For the first 48 odd numbers:n = 48a = 1d = 2→ sum = 48/2 (2 x 1 + (48 - 1) x 2)= 482= 2304The sum of the first n odd numbers is always n2
These are easy if you work this way... 72 - 48 = 24, half of 24 is 12 which is the smaller number and 12 + 48 = 60which is the other number.
If the two numbers have a hcf of 48 then both numbers must be multiples of 48. 384 ÷ 48 = 8 .........Taking the paired factors of 48 of both numbers then the sum of the other factors = 8 but there can be no other common factor. 1 + 7 = 8, 3 + 5 = 8 are solutions but 2 + 6 = 8 and 4 + 4 = 8 are not, as further common factors of 2 and 4 respectively are present which would increase the hcf to 96 and 192. The numbers are 1 x 48 and 7 x 48 → 48 and 336 and.....................3 x 48 and 5 x 48 → 144 and 240
sum = 48
46
They are: -8 and 6
16 AND 12
2 + 4 + 6 + ... + 44 + 46 + 48 = 600
30 + 18 = 48 30 - 18 = 12 The two numbers are therefore 30 and 18.
x+ (x+1) + (x+2) = 48 3x+3 = 48 3x = 45 x = 15
1/a + 1/b = b/ab + a/ab = (b+a)/ab So, the answer to your question is 32/48 or 2/3
336 and 48
the sum of two whole numbers are 77 .their difference is 48.what are the two numbers
2304Sum of an arithmetic progression = n/2 (2a + (n-1)d) where n is the number of items, a is the first item and d is the difference between them:For the first 48 odd numbers:n = 48a = 1d = 2→ sum = 48/2 (2 x 1 + (48 - 1) x 2)= 482= 2304The sum of the first n odd numbers is always n2
Numbers are 47 and 48