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The sum of first seven terms of an A.P. is $182$. If its $4^{\text{th}}$ term and the $17^{\text{th}}$ term are in the ratio $1: 5$, find the A.P.
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Let $a$ be the first term and $d$ be the common difference.
$S_7 = 182 \Rightarrow \frac{7}{2}(2a + 6d) = 182$
$\Rightarrow 2a + 6d = \frac{182 \times 2}{7} = 52$
$a + 3d = 26$
quad ----- (i)
$\frac{a_4}{a_{17}} = \frac{1}{5} \Rightarrow \frac{a + 3d}{a + 16d} = \frac{1}{5}$
$\Rightarrow 5a + 15d = a+16d$
$4a = d$
quad --------(ii)
Solving (i) and (ii)
$a = 2$ and $d=8$
$\therefore \text{AP is } 2, 10, 18, 26, ....$
$S_7 = 182 \Rightarrow \frac{7}{2}(2a + 6d) = 182$
$\Rightarrow 2a + 6d = \frac{182 \times 2}{7} = 52$
$a + 3d = 26$
quad ----- (i)
$\frac{a_4}{a_{17}} = \frac{1}{5} \Rightarrow \frac{a + 3d}{a + 16d} = \frac{1}{5}$
$\Rightarrow 5a + 15d = a+16d$
$4a = d$
quad --------(ii)
Solving (i) and (ii)
$a = 2$ and $d=8$
$\therefore \text{AP is } 2, 10, 18, 26, ....$