Geometric progression

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

Correct answer:

n =  3

Step-by-step explanation:

$$\begin{gathered} a_1=7 \\ q=5 \\ s=217 \\ s=aq\cdot {\displaystyle \frac{q^n-1}{q-1}} \\ s\mathrm{/}{a}_1\cdot (q-1)=q^n-1 \\ s\mathrm{/}{a}_1\cdot (q-1)+1=q^n \\ \ln s\mathrm{/}{a}_1\cdot (q-1)+1=n\ln q \\ n={\displaystyle \frac{\ln (s\mathrm{/}{a}_1\cdot (q-1)+1)}{\ln (q)}}={\displaystyle \frac{\ln (217\mathrm{/}7\cdot (5-1)+1)}{\ln (5)}}=3 \end{gathered}$$

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