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Question
In an AP given d = 5, S9 = 75, find a and a9.
Solution
Here, d = 5 and S9 = 75 (given)
∵ Sn = `"n"/2` [a + an]
⇒ `S_9 = 9/2[a + a_9] = 75`
⇒ 3a + 3a9 = 50 ...(1)
∵ `S_n = n/2[2a + (n - 1) xx d]`
⇒ `S_9 = 9/2[2a + (9 - 1) xx 5] = 75`
⇒ `9/2[2a + 40] = 75`
⇒ 9a + 180 = 75
⇒ 9a = 75 - 180
⇒ 9a = -105
⇒ a = `-105/9`
⇒ a = `-35/3` ...(2)
By substituting the value of a from equation (2) in equation (1),
`3 xx (-35/3) + 3a_9 = 50`
⇒ -35 + 3a9 = 50
⇒ 3a9 = 50 + 35
⇒ 3a9 = 85
⇒ a9 = `85/3`
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