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Question
The 104th term and 4th term of an A.P. are 125 and 0. Find the sum of first 35 terms
Solution
t104 = 125
t4 = 0
a + (n – 1)d = tn
a + 103d = 125 ...(1)
a + 3d = 0 ...(2)
(–) (–) (–)
(1) – (2) ⇒ 100d = 125
d = `125/100 = 5/4`
Substitute d = `5/4` in (2)
`"a" + 3 xx 5/4` = 0
`"a" + 15/4` = 0
⇒ a = `-15/4`
∴ Sn = `"n"/2 (2"a" + ("n" - 1)"d")`
S35 = `35/2(2 xx (-15)/4 + 34 xx 5/4)`
= `35/2((-15)/2 + 85/2)`
= `35/2(70/2) = 35/2 xx 35`
= `1225/2`
= 612.5
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