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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.

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by mits

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by mits

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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

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