1/5 + 3/5 = 4/5
5 - 5/3 + 5/9 - 5/27 + ... = 5 + 5(-1/3)¹ + 5(-1/3)² + 5(-1/3)³ + ... The required sum is an infinite GP with initial term a = 5, and common difference r = -1/3 As |r| < 1, the sum can be found from sum = a/(1 - r) → 5 - 5/3 + 5/9 - 5/27 + ... = 5/(1 - (-1/3)) = 5/(1 + 1/3) = 5/(4/3) = 5 × 3/4 = 15/4 = 3¾
The sum of -1/15 and -3/5 is -2/3
The sum of 5/6 and 5/6 is 1 2/3
1.5 or 1 + 1 over 2
1/5 + 3/5 = 4/5
the sum would be 1/4 + 3/5 = 5/20 + 12/20 = 17/20
5 - 5/3 + 5/9 - 5/27 + ... = 5 + 5(-1/3)¹ + 5(-1/3)² + 5(-1/3)³ + ... The required sum is an infinite GP with initial term a = 5, and common difference r = -1/3 As |r| < 1, the sum can be found from sum = a/(1 - r) → 5 - 5/3 + 5/9 - 5/27 + ... = 5/(1 - (-1/3)) = 5/(1 + 1/3) = 5/(4/3) = 5 × 3/4 = 15/4 = 3¾
The sum of -1/15 and -3/5 is -2/3
The sum of 5/6 and 5/6 is 1 2/3
1.5 or 1 + 1 over 2
The sum of -3/5 and 1/7 is -16/35
The sum of 5/6 and 3/8 = 1 5/24
The sum of 3/8 and 1/4 is 5/8.
3/5 + 1/2 = 6/10 + 5/10 = 11/10 = 11/10
The sum of 4/5 and 4/5 is 1 3/5
2/9 + 1/3 = 2/9 + 3/9 - Note the equivalence of the fraction 1/3 & 3/9 Add the numerators 5/9 The Answer . 5/9 = 0.555.... ( recurring to infinity as a decimal).