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how many term has G.p whose 2nd term is 1/2 and common ratio and the last term are 1/4and1/128respestively

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The geometric progression has the form:


\mleft\lbrace a,ar,ar^2,ar^3,\ldots,ar^n\mright\rbrace

We have the information about the second term, a*r:


ar=(1)/(2)

We know that the common ratio is


r=(1)/(4)

So from this information we can get the coefficient a:


\begin{gathered} ar=(1)/(2) \\ a\cdot(1)/(4)=(1)/(2) \\ a=(4)/(2)=2 \end{gathered}

And we also know that the last term is 1/128, that is


ar^n=(1)/(128)

From this one we can find n:


\begin{gathered} 2\cdot((1)/(4))^n=(1)/(128) \\ ((1)/(4))^n=(1)/(128\cdot2) \end{gathered}

We can apply the property of the logarithm of power to get n:


\begin{gathered} \log (((1)/(4))^n)=\log ((1)/(256)) \\ n\cdot\log ((1)/(4))^{}=\log ((1)/(256)) \\ n=(\log ((1)/(256)))/(\log ((1)/(4))) \\ n=4 \end{gathered}

Be careful, because n is not the number of terms. The number of terms is n+1, so the G.P. has 5 terms

User Bruce Afruz
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