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Question
In an A.P. (with usual notations) : given d = 5, S9 = 75, find a and a9
Solution
d = 5, S9 = 75
an = a + (n – 1)d
a9 = a + (9 – 1) x 5
= a + 40 ...(i)
S9 = `n/(2)[2a + (n - 1)d]`
75 = `(9)/(2)[2a + 8 xx 5]`
`(150)/(9)` = 2a + 40
2a = `(150)/(9) - 40`
= `(50)/(3) - 40`
2a = `(-70)/(3)`
⇒ a = `(-70)/(2 xx 3)`
a = `(-35)/(3)`
From (i),
a9 = a + 40
= `(-35)/(3) + 40`
= `(-35 + 120)/(3)`
= `(85)/(3)`
∴ a = `(-35)/(3), a_9 = (85)/(3)`.
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