Advertisements
Advertisements
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.
Sum
Advertisements
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
shaalaa.com
Is there an error in this question or solution?
