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Question
Find t21, if S41 = 4510 in an A.P.
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Solution
For an A.P., let a be the first term and d be the common difference.
S41 = 4510 ...[Given]
Since `S_n = n/2 [2a + (n - 1)d]`,
`S_41 = 41/2 [2a + (41 - 1)d]`
∴ `4510 = 41/2 (2a + 40d)`
∴ `4510 = 41/2 xx 2 (a + 20d)`
∴ 4510 = 41(a + 20d)
∴ a + 20d = `4510/41`
∴ a + 20d = 110 ...(i)
Now, tn = a + (n – 1)d
∴ t21 = a + (21 – 1)d
= a + 20d
∴ t21 = 110 ...[From (i)]
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