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Question

In an AP, given d$$=5$$, $$S_9=75$$, find a and $$a_9$$.


Solution

$$d=5,s_9=75$$
$${ s }_{ 9 }=\cfrac { 9 }{ 2 } [2a+(9-1)5]\\ 75=\cfrac { 9 }{ 2 } [2a+40]\\ 150=9[2a+40]\\ 150=18a+360\\ 150-360=18a\\ -210=18a\\ \cfrac { -210 }{ 18 } =a\\ a=\cfrac { -35 }{ 3 } \\ { a }_{ 9 }=a+(9-1)5\\ { a }_{ 9 }=\cfrac { -35 }{ 3 } +40\\ { a }_{ 9 }=\cfrac { -35+120 }{ 3 } \\ { a }_{ 9 }=\cfrac { -85 }{ 3 } $$

Mathematics

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