Calculate 81860

The two terms of the geometric sequence are a2=12 and a5=three halves.
a) calculate the tenth term of the sequence.
b) calculate the sum of the first 8 terms of the sequence.
v) how many first terms of the sequence need to be added so that the sum is equal to 45?

Correct answer:

a₁₀ =  3/64
s₈ =  47.8125
n =  4

Step-by-step explanation:

$$\begin{gathered} a_2=12 \\ {a}_5=3/2=\frac{3}{2}=1\frac{1}{2}=1.5 \\ {a}_5=a_2\cdot q^3 \\ q=\sqrt[3]{a_5/a_2}=\sqrt[3]{1.5/12}=\frac{1}{2}=0.5 \\ {a}_{10}=a_5\cdot q^5=1.5\cdot 0.5^5=\frac{3}{64}=0.0469 \end{gathered}$$

$$\begin{gathered} a_1=a_2/q=12/0.5=24 \\ {s}_8=a_1\cdot \frac{q^8-1}{q-1}=24\cdot \frac{0.5^8-1}{0.5-1}=\frac{765}{16}=47\frac{13}{16}=47.8125 \end{gathered}$$

$$\begin{gathered} s=45 \\ s=a_1\cdot \frac{q^n-1}{q-1} \\ s/a_1\cdot (q-1)=q^n-1 \\ C=s/a_1\cdot (q-1)+1=45/24\cdot (0.5-1)+1=\frac{1}{16}=0.0625 \\ {q}^n=C \\ n\ln q=\ln C \\ n=\ln (C)/\ln q=\ln 0.0625/\ln 0.5=4 \end{gathered}$$

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