# Sequences 2 - Solutions

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

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