In an AP, given d $= 5$ , $S_{9} = 75$ , find a and $a_{9}$ .

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Solution

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

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In an AP, given d=5, S9=75, find a and a9.

d=5,s9=75s9=92[2a+(9−1)5]75=92[2a+40]150=9[2a+40]150=18a+360150−360=18a−210=18a−21018=aa=−353a9=a+(9−1)5a9=−353+40a9=−35+1203a9=−853

In an AP, given d $= 5$ , $S_{9} = 75$ , find a and $a_{9}$ .

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