If $5^{t h}$ and $6^{t h}$ terms of an A.P. are respectively $6$ and $5$ , find the $11^{t h}$ terms of an A.P.

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Solution

$T_{n} = a + (n - 1) d$

$T_{5} = a + (5 - 1) d = a + 4 d$

$\Rightarrow T_{5} = a + 4 d = 6$ ...(1)

Similarly,

$T_{6} = a + 5 d = 5$ ...(2)

(2)-(1) gives

$5 d - 4 d = 5 - 6 \Rightarrow d = - 1$

From (1)

$a + 4 d = 6 \Rightarrow a = 6 - 4 (- 1) = 10$

Now,

$T_{11} = 10 + (11 - 1) (- 1)$

$= 10 - 10$

$= 0$

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