no.9
this question deals with progression and it's a simple example of GP
to solve this question, let's find two things
1. first term
2. common difference
first term = 2
common ratio = 2
[tex]\begin{gathered} T_7=ar^6 \\ T_7=2\times2^6 \\ T_7=2\times64 \\ T_7=128 \\ \\ T_8=ar^7 \\ T_8=2\times2^7 \\ T_8=2\times128 \\ T_8=256 \end{gathered}[/tex]10. this question relates to arithmetic progression
[tex]\begin{gathered} first\text{ term = 66} \\ \text{common difference = 16} \\ T_n=a+(n-1)d \\ T_5=66+(5-1)\times16 \\ T_5=66+4\times16 \\ T_5=66+64 \\ T_5=130 \\ \\ T_6=66+(6-1)\times16 \\ T_6=66+5\times16 \\ T_6=66+80 \\ T_6=146 \\ \\ T_7=66+(7-1)\times16 \\ T_7=66+6\times16 \\ T_7=66+96 \\ T_7=162 \\ \\ T_8=66+(8-1)\times16 \\ T_8=66+7\times16 \\ T_8=66+112 \\ T_8=178 \end{gathered}[/tex]11. This question relates to geometric progression
first term(a) = 160
common ratio (r) = 1/2
[tex]\begin{gathered} T_n=ar^{n-1} \\ T_5=160\times(\frac{1}{2})^{5-1} \\ T_5=160\times(0.5)^4 \\ T_5=160\times\frac{1}{16} \\ T_5=10 \\ \\ T_6=ar^{6-1} \\ T_6=160\times(\frac{1}{2})^5 \\ T_6=160\times\frac{1}{32} \\ T_6=5 \\ \\ T_7=ar^{7-1} \\ T_7=ar^6 \\ T_7=160\times(\frac{1}{2})^6 \\ T_7=160\times\frac{1}{64} \\ T_7=\frac{5}{2}=2.5 \\ \\ \\ T_8=ar^{8-1} \\ T_8=ar^7 \\ T_8=160\times(\frac{1}{2})^7 \\ T_8=160\times\frac{1}{128} \\ T_8=\text{ }\frac{5}{4}=1.25 \end{gathered}[/tex]12. This question relates to arithmetic progression
first term (a) = -9
common difference(d) = 7
[tex]\begin{gathered} T_n\text{ = a+(n-1)d} \\ T_5=a+(5-1)d \\ T_5=a+4d \\ T_5=-9+(4\times7) \\ T_5=-9+28 \\ T_5=19 \\ \\ T_6=a+(6-1)d \\ T_6=a+5d \\ T_6=-9+(5\times7) \\ T_6=-9+45 \\ T_6=36 \\ \\ T_7=a+(7-1)d \\ T_7=a+6d \\ T_7=-9+(6\times7) \\ T_7=-9+42 \\ T_7=33 \\ \\ T_8=a+(8-1)d \\ T_8=a+7d \\ T_8=-9+(7\times7) \\ T_8=-9+49 \\ T_8=40 \end{gathered}[/tex]