Question

Please help with both questions thank you

Answer 1

n

∑(.2k +.3)

k=1

n =15

Explanation:

The initial value is .5

The common difference is .2 (That is what we add each time)

an = a1 + d(n-1)

an = .5 +.2(n-1)

an = .5 +.2n-.2

an = .2n +.3

We need to sum this up for n times. Replace n in the function with k and sum up to n

n

∑(.2k +.3)

k=1

Now we want to find the sum and see when it is greater than 26.2

Given the formula for Sn

Sn = n/2 (2a1 + d(n-1))

26.2 = n/2 (2*.5 +.2 (n-1))

Multiply each side by 2

52.4 = n (1+.2n-.2)

52.4 = n(.8+.2n)

52.4 = .8n +.2n^2

Subtract 52.4 from each side

0 = .2 n^2 +.8n -52.4

Using the quadratic equations

b^2 -4ac

-------------

2a

n≈14.3095

Rounding to the next highest number since we need to be greater than 26.2

n=15