Geometric sequence 3

In geometric sequence is a₈ = 312500; a₁₁= 39062500; s_n=1953124. Calculate the first item a₁, quotient q, and n - number of members by their sum s_n.

Correct answer:

a₁ =  4
q =  5
n =  9

Step-by-step explanation:

$$\begin{gathered} a_8=a_1{q}^7=312500 \\ {a}_{11}=a_1{q}^{10}=39062500 \\ {\displaystyle \frac{312500}{39062500}}={\displaystyle \frac{q^7}{q^{10}}} \\ 0.008=q^{-3} \\ 0.008^{-0.333333333333}=q \\ q=5 \\ {a}_8=a_1{q}^7=a_1\cdot 5^7=312500 \\ {a}_1={\displaystyle \frac{312500}{5^7}}=4 \end{gathered}$$

$$\begin{gathered} {s}_n=1953124=a_1{\displaystyle \frac{q^n-1}{q-1}}={\displaystyle \frac{5^n-1}{5-1}} \\ 1953124=5^n-1 \\ 1953125=5^n \\ \ln 1953125=n\ln 5 \\ n={\displaystyle \frac{\ln 1953125}{\ln 5}} \\ n=9 \end{gathered}$$

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