This page offers solutions to the set of IB style questionsThe first 4 terms of an arithmetic sequence are shown below3, 9, 15, 21Solution\(\begin{align} & a)\quad 27\quad (Common\quad difference\quad is\quad 6)\\ & b)\quad { U }_{ n }={ U }_{ 1 }+d(n-1),\quad So\quad { U }_{ n }=3+6(n-1)\\& \quad \quad \quad So\quad { U }_{ 100 }=597\\ & c)\quad { S }_{ n }=\frac { n }{ 2 } ({ 2U }_{ 1 }+d(n-1))\\& \quad \quad \quad { S }_{ 30 }=\frac { 30 }{ 2 } (2\times 3+6(30-1))\\& \quad \quad \quad { S }_{ 30 }=2700 \end{align}\)A geometric Sequence has all its terms positive. The first term is 9 and the third term is 144.Solution\(\begin{align} & a)\quad { U }_{ 1 }\times { r }^{ 2 }={ U }_{ 3 }\\& 9\times { r }^{ 2 }=144\\& { r }^{ 2 }=\frac { 144 }{ 9 } \\& r=4\quad \quad (since\quad all\quad terms\quad are\quad positive,\quad r\quad =\quad -4\quad is\quad not\quad possible)\\ & b)\quad { S }_{ n }={ U }_{ 1 }\times \frac { ({ r }^{ n }-1) }{ r-1 } \\& So\quad { S }_{ n }=9\times \frac { ({ 4 }^{ 10 }-1) }{ 4-1 } \\& { S }_{ n }=3145725 \end{align} \)The 4th term of an arithmetic sequence is 30 and the 11th term is 51.Solution\(\begin{align} & a)\quad { U }_{ 4 }+7d={ U }_{ 11 }\\& So,\quad 30+7d=51\\& 7d=21\quad and\quad d=3\\ & b)\quad { U }_{ 4 }-3d={ U }_{ 1 }\\& So,\quad 30-3\times 3={ U }_{ 1 }=21\\ & c)\quad { U }_{ n }={ U }_{ 1 }+d(n-1)\\& So,\quad { U }_{ 42 }=21+3(42-1)\\& { U }_{ 42 }=144\\ & d)\quad { S }_{ n }=\frac { n }{ 2 } ({ 2U }_{ 1 }+d(n-1))\\& So,\quad { S }_{ 150 }=\frac { 150 }{ 2 } (2\times 21+3(150-1))\\& { S }_{ 150 }=36675 \end{align}\)The population of Whoville is growing each year. At the end of 2004 the population...