Question

Find the following sum 1. 1+2+3...........+100 2. 3+7+11.........+199

Answer 1

Answer:

Use Sn = $\frac{n}{2} (a + An)$

Step-by-step explanation:

First Question is easy as n is 100 as numbers are in counting order

Sn = $\frac{100}{2}( 1 + 100)\\= 50 *51\\=2550$

In Second question, we need to find n first by using An = a + (n-1)d

d = 7 - 3 = 4

199 = 3 + (n-1)4

199 = 3 + 4n - 4

199 + 1 = 4n

n = 200/4

n = 50

Now use Sn formulae

Sn = $\frac{50}{2}( 3 + 199)\\ = 25 * 202\\= 5050$