An A. P. consists of 57 terms of which $7^{t h}$ term is 13 and the last term is 108. Find the $45^{t h}$ term of this A. P.

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Solution

Let first term = a,

common difference = d

$a_{7} = 13 a + 6 d = 13 - - - - (1) a_{57} = 108 a + 56 d = 108 - - - - (2)$

(2) - (1)

$50 d = 95 d = \frac{95}{50} = 1.9$

Now from (1),

$a + 6 (1.9) = 13 a = 13 - 11.4 = 1.6$

$a_{45} = a + 44 d = 1.6 + 44 \times 1.9 = 85.2$

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