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Question
The sum of first seven terms of an A.P. is 182. If its 4th and the 17th terms are in the ratio 1 : 5, find the A.P.
Solution
Let a be the first term and d be the common difference.
We know that, sum of first n terms = Sn = \[\frac{n}{2}\][2a + (n − 1)d]
According to the question,
\[S_7 = 182\]
\[\Rightarrow \frac{7}{2}\left[ 2a + \left( 7 - 1 \right)d \right] = 182\]
\[\Rightarrow \frac{1}{2}\left( 2a + 6d \right) = 26\]
\[\Rightarrow a + 3d = 26\]
\[\Rightarrow a = 26 - 3d ....(1)\]
Also,
\[\frac{a_4}{a_{17}} = \frac{1}{5}\]
\[\Rightarrow \frac{a + (4 - 1)d}{a + (17 - 1)d} = \frac{1}{5}\]
\[\Rightarrow \frac{a + 3d}{a + 16d} = \frac{1}{5}\]
\[\Rightarrow 5(a + 3d) = a + 16d\]
\[\Rightarrow 5a + 15d = a + 16d\]
\[\Rightarrow 5a - a = 16d - 15d\]
\[\Rightarrow 4a = d ....(2)\]
On substituting (2) in (1), we get
\[a = 26 - 3\left( 4a \right)\]
\[\Rightarrow a = 26 - 12a\]
\[\Rightarrow 12a + a = 26\]
\[\Rightarrow 13a = 26\]
\[\Rightarrow a = 2\]
\[\Rightarrow d = 4 \times 2 \left[ \text{ From } \left( 2 \right) \right]\]
\[ \Rightarrow d = 8\]
Thus, the A.P. is 2, 10, 18, 26, ......
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