Question

1. Find Sn for the arithmetic series where a1= 3 , an= 42 , n= 14

2. Find sn for the arithmetic series where a1= −20 , n= 25 , an= 148
3. Find sn for the arithmetic series where a1= 5 , d= 1/2 , n= 13

Answer 1

1. Sn = n/2 * (a1 + an) = 14/2 * (3 + 42) = 14/2 * 45 = 315

2. Sn = n/2 * (a1 + an) = 25/2 * (-20 + 148) = 25/2 * (128) = 6400

3. Sn = n/2 * (2a1 + (n-1)d) = 13/2 * (2 * 5 + (13-1) * (1/2)) = 13/2 * (10 + 6 * (1/2)) = 13/2 * (10 + 3) = 13/2 * 13 = 169

Answer 2

Sn for an arithmetic series can be found using the formula: Sn = (n/2) * (a1 + an). In this case, we have:

a1 = 3, an = 42, n = 14

Sn = (14/2) * (3 + 42) = (14/2) * 45 = 315

Sn for an arithmetic series can be found using the formula: Sn = (n/2) * (a1 + an). In this case, we have:

a1 = -20, an = 148, n = 25

Sn = (25/2) * (-20 + 148) = (25/2) * 64 = 800

Sn for an arithmetic series can be found using the formula: Sn = n/2 * (2a1 + (n-1)d) where d is the common difference. In this case, we have:

a1 = 5, n = 13, d = 1/2

Sn = 13/2 * (2 * 5 + (13 - 1) * 1/2) = 13/2 * (10 + 12/2) = 13/2 * (10 + 6) = 13/2 * 16 = 104

So for the series with the given parameters, Sn = 104.