The sum of first seven terms of an A.P. is 182. If its 4^{th} and the 17^{th} 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 = *S*_{n }= \[\frac{n}{2}\][2*a* + (*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, ..... .