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)=3a_{1}=7 \ \\ q=5 \ \\ s=217 \ \\ \ \\ s=aq \cdot \ \dfrac{ q^n-1 }{ q-1 } \ \\ s/a_{1} \cdot \ (q-1)=q^n-1 \ \\ s/a_{1} \cdot \ (q-1)+1=q^n \ \\ \ln s/a_{1} \cdot \ (q-1)+1=n \ln q \ \\ \ \\ n=\dfrac{ \ln( s/a_{1} \cdot \ (q-1)+1) }{ \ln(q) }=\dfrac{ \ln( 217/7 \cdot \ (5-1)+1) }{ \ln(5) }=3

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