# In a arithemetic progression the first term is 1 and the last term is 31. if eighth term ::third from last term =5:9. find n.

1
by mits

2014-07-23T17:13:45+05:30

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Tn = T1 + (n-1) a
T1 = 1          Tn  =  31            Tn - T1 = 30  =>  (n-1) a = 30    --- equation 1
T8 = T1 + 7 a       T(n-2) = T1 + (n-3) a
T8 / T(n-2)  = 5 / 9  =>  T(n-2) / T8 = 9/5              subtract both sides from 1,
[  T(n-2) - T8  ]  / T8  =  4 / 5            =>  (n - 10) a / (1 +7a) = 4 / 5
=> 5 a (n - 10)  = 4 + 28 a    => 5 an - 78 a - 4  =0     ---- equation 2
From equation 1 ,   na = a + 30  , substitute in eq 2

5 (a+30) - 78a - 4 =0    => - 73 a + 146 = 0  => a = 2
from equaion 1 , we get     n -1 = 15 => n = 16