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
Numbers are 47 and 48
You can either do this by trial-and error, or solve the equation n + (n + 2) + (n + 4) = 48.