2 + 3 - 4 + 5 - ____+ 99 - 100 + 101 will be equal to

123 100  48,049,831 Forty-eight million forty-nine thousand eight hundred thirty one EXERCISE 1 7,707 tickets 2 3,020 runs 3 2,28,800 votes  100th term of the arithmetic sequence 1, 2, 3, = 100 b) The position of 100 in the arithmetic sequence 2, 4, 6, = 50

Since the only operation inside the brackets is a single sum, we first add 1 1 to 100 100 to get 101 101 4×3−1 2 + 4 × 3 − 1 Solution Applying BODMAS, we 100 multiples of 5 greater than 100 are: 105, 110, 115, 120 : 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19

Solution By Steps Step 1: Calculate 1! Step 2: Calculate 2! Step 3: Calculate 3! Step 4: Calculate 4! The pattern observed is that each It is an Arithmetic progression with common difference 1 Therefore sum is 2 * ie * = 5050