If the 5th term and the 14th term of an AP are 35 and 8 respectively, then find the 20th term of the AP.

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Solution

The correct option is D $- 10$
Let $a$ be the first term and $d$ be their common difference of the AP.
Then, $n t h$ term of the AP $T_{n} = a + (n - 1) d$
Given, $T_{5} = 35$
$=> a + (5 - 1) d = 35$
$=> a + 4 d = 35$ -- (1)
$T_{1} 4 = 8$
$=> a + (14 - 1) d = 8$
$=> a + 13 d = 8$ -- (2)
Subtracting eqn 2 from eqn 1, we get
$9 d = - 27$
$=> d = - 3$
So, $a + 4 (- 3) = 35$
$=> a = 47$
So, $T_{20} = a + (20 - 1) d = 47 + 19 (- 3) = - 10$

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