Geometric progression

In geometric progression, a1 = 7, q = 5. Find the condition for n to sum first n members is: sn≤217.

Correct result:

n =  3

Solution:

a1=7 q=5 s=217  s=aq⋅ qn−1q−1 s/a1⋅ (q−1)=qn−1 s/a1⋅ (q−1)+1=qn ln⁡s/a1⋅ (q−1)+1=nln⁡q  n=ln⁡(s/a1⋅ (q−1)+1)ln⁡(q)=ln⁡(217/7⋅ (5−1)+1)ln⁡(5)=3

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