Answer:
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