# Find t21, if S41 = 4510 in an A.P. - Algebra

Sum

Find t21, if S41 = 4510 in an A.P.

#### Solution

For an A.P., let a be the first term and d be the common difference.

S41 = 4510     ......[Given]

Since Sn = "n"/2 [2"a" + ("n" - 1)"d"],

S41 = 41/2 [2"a" + (41 - 1)"d"]

∴ 4510 = 41/2 (2"a" + 40"d")

∴ 4510 = 41/2 xx 2 ("a" + 20"d")

∴ 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)]

Concept: Sum of First n Terms of an AP
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