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Question
An A.P. consists of 57 terms of which 7th term is 13 and the last term is 108. Find the 45th term of this A.P.
Solution
Number of term = n = 57
t7 = 13
`=>` a + 6d = 13 ...(i)
Last term = t57 = 108
`=>` a + 56d = 108 ...(ii)
Subtracting (i) from (ii), we get
50d = 95
`=> d = 95/50`
`=> d = 19/10`
Substituting value of d in (i), we get
`a + 6 xx 19/10 = 13`
`=> a + 57/5 = 13`
`=> a = 13 - 57/5`
= `(65 - 57)/5`
= `8/5`
`=>` General term = tn
= `8/5 + (n - 1) xx 19/10`
`=> t_45 = 8/5 + 44 xx 19/10`
= `8/5 + 418/5`
= `426/5`
= 85.2
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