1.

the 17th term of an ap is 5 more than twice of its 8th term and the 11th term of the ap is 43 then the nth termis

Answer»

T11 = 43

=> a + 10d = 43 -------(1)

T17 = 2× T8 + 5

=> a + 16d = 2 (a + 7d) + 5

=> a + 16d = 2a + 14D + 5

=> - a + 2d = 5

=> 2d - a = 5 -------(2)

On adding equation 1 and 2, we get

12d = 48

=> d = 4

Now,

On SUBSTITUTING the value d in equation 2, we get

2 (4) - a = 5

=> 8 - a = 5

=> a = 3

Tn = a + (n-1)d

= 3 + ( n-1)(4)

= 3 + 4N - 4

= 4n - 1


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