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Question
Find the sum of all the 11 terms of an AP whose middle most term is 30.
Solution
Since, the total number of terms (n) = 11 ...[odd]
∴ Middle most term = `(n + 1)^("th")/2` term
= `((11 + 1)/2)^("th")` term
= 6th term
Given that,
a6 = 30 ...[∵ an = a + (n − 1)d]
⇒ a + (6 − 1)d = 30
⇒ a + 5d = 30 ...(i)
∵ Sum of n terms of an AP,
Sn = `n/2[2a + (n - 1)d]`
∴ S11 = `11/2[2a + (11 - 1)d]`
= `11/2(2a + 10d)`
= 11(a + 5d)
= 11 × 30 ...[From equation (i)]
= 330
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