Question

\[ \begin{array}{lr} & \\ & =16 \times \frac{1}{16} \\ & =1 \\ r_{6} & =16\left(\frac{1}{2}\right)^{6-1} \\ & =16\left(\frac{1}{2}\right)^{5} \\ & =16 \times \frac{1}{32} \\ & =\frac{1}{2} \\ & \text { Find } a \\ & \\ \text { spectively } & a=\frac{1}{2} \\ & a=\frac{1}{4 r^{2}} \\ & \\ & a=\frac{1}{4\left(\frac{1}{2}\right)^{2}} \\ & a=\frac{1}{4 \times \frac{1}{4}} \end{array} \] is positive) Assigment The \( 8^{\text {th }} \) term of a G.p is 640 if the first of Gip is 5 Find the (a) Comment ratio (b) \( 10^{\text {th }} \) term ECampl are 6 of the \[ \begin{array}{l} \underline{b}= \\ T_{2}= \\ \text { ar2 } \end{array} \] \( a r^{\prime}= \) di

          \[
\begin{array}{lr} 
& \\
& =16 \times \frac{1}{16} \\
& =1 \\
r_{6} & =16\left(\frac{1}{2}\right)^{6-1} \\
& =16\left(\frac{1}{2}\right)^{5} \\
& =16 \times \frac{1}{32} \\
& =\frac{1}{2} \\
& \text { Find } a \\
& \\
\text { spectively } & a=\frac{1}{2} \\
& a=\frac{1}{4 r^{2}} \\
& \\
& a=\frac{1}{4\left(\frac{1}{2}\right)^{2}} \\
& a=\frac{1}{4 \times \frac{1}{4}}
\end{array}
\]

is positive)

Assigment
The \( 8^{\text {th }} \) term of a G.p is 640 if the first of Gip is 5 Find the (a) Comment ratio (b) \( 10^{\text {th }} \) term

ECampl are 6 of the
\[
\begin{array}{l}
\underline{b}= \\
T_{2}= \\
\text { ar2 }
\end{array}
\]
\( a r^{\prime}= \)

di

=16 ×(1)/(16)
=1

r6 =16((1)/(2))^6-1
=16((1)/(2))^5
=16 ×(1)/(32)
=(1)/(2)
Find a

spectively a=(1)/(2)
a=(1)/(4 r^2)

a=(1)/(4((1)/(2))^2)
a=(1)/(4 ×(1)/(4))

is positive)

Assigment
The 8^th term of a G.p is 640 if the first of Gip is 5 Find the (a) Comment ratio (b) 10^th term

ECampl are 6 of the

b=

T2=
ar2

a r^'=

di

Added by Stephen C.

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