In an Arithmetic progression the first term is $2$ and the sum of the first five terms is one fourth of next five terms. Show that the $20^{t h}$ term is equal to $- 112$ .

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Solution

In an Arithmetic Progression $a = 2$
$a + (a + d) + (a + 2 d) + (a + 3 d) + (a + 4 d) = \frac{1}{4} [(a + 5 d) + (a + 6 d) + (a + 7 d) + (a + 8 d) + (a + 9 d)]$
$∴$ $5 a + 10 d = \frac{1}{4} (5 a + 35 d)$
Substitute $a = 2$ , we get
$\Rightarrow 5 \times 2 + 10 d = \frac{1}{4} (5 \times 2 + 35 d)$
$\Rightarrow 10 + 10 d = \frac{1}{4} (10 + 35 d)$
$\Rightarrow 40 + 40 d = 10 + 35 d$
$\Rightarrow$ $40 d - 35 d = 10 - 40$
$\Rightarrow 5 d = - 30$
$\Rightarrow$ $d = \frac{- 30}{5} = - 6$
$\Rightarrow$ $d = - 6$
$20^{t h}$ term in A.P. $= a + 19 d$
$= 2 + 19 (- 6)$
$= 2 + (- 114)$
$= - 112$
$∴$ $T_{20} = - 112$

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