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Question
The sum of first 7 terms of an A.P. is 10 and that of next 7 terms is 17. Find the progression.
Solution
\[\text { We have: } \]
\[ S_7 = 10\]
\[ \Rightarrow \frac{7}{2}\left[ 2a + (7 - 1)d \right] = 10\]
\[ \Rightarrow \frac{7}{2}\left[ 2a + 6d \right] = 10\]
\[ \Rightarrow a + 3d = \frac{10}{7} . . . (i)\]
\[\text { Also, the sum of the next seven terms } = S_{14} - S_7 = 17\]
\[ \Rightarrow \frac{14}{2}\left[ 2a + \left( 14 - 1 \right)d \right] - \frac{7}{2}\left[ 2a + (7 - 1)d \right] = 17\]
\[ \Rightarrow 7\left[ 2a + 13d \right]\]
\[ - \frac{7}{2}\left[ 2a + 6d \right] = 17\]
\[ \Rightarrow 14a + 91d - 7a - 21d = 17\]
\[ \Rightarrow 7a + 70d = 17\]
\[ \Rightarrow a + 10d = \frac{17}{7} . . . (ii)\]
\[\text { From (i) and (ii), we get }: \]
\[\frac{10}{7} - 3d = \frac{17}{7} - 10d\]
\[ \Rightarrow 7d = 1\]
\[ \Rightarrow d = \frac{1}{7}\]
\[\text { Putting the value in (i), we get: } \]
\[a + 3d = \frac{10}{7}\]
\[ \Rightarrow a + \frac{3}{7} = \frac{10}{7}\]
\[ \Rightarrow a = 1\]
\[ \therefore a = 1, d = \frac{1}{7}\]
The progression thus formed is
\[1, \frac{8}{7}, \frac{9}{7}, \frac{10}{7} . . .\]
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