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Question
The 4th term of an A.P. is three times the first and the 7th term exceeds twice the third term by 1. Find the first term and the common difference.
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Solution
Given:
Let the first term of the A.P. be a and the common difference be d.
\[a_4 = 3a\]
\[ \Rightarrow a + \left( 4 - 1 \right)d = 3a\]
\[ \Rightarrow a + 3d = 3a\]
\[ \Rightarrow 3d = 2a\]
\[ \Rightarrow a = \frac{3d}{2} . . . (i)\]
\[\text { And,} a_7 - 2 a_3 = 1\]
\[ \Rightarrow a + \left( 7 - 1 \right)d - 2\left[ a + \left( 3 - 1 \right)d \right] = 1\]
\[ \Rightarrow a + 6d - 2(a + 2d) = 1\]
\[ \Rightarrow a + 6d - 2a - 4d = 1\]
\[ \Rightarrow - a + 2d = 1\]
\[ \Rightarrow - \frac{3d}{2} + 2d = 1 \left[ \text { From } (i) \right] \]
\[ \Rightarrow \frac{- 3d + 4d}{2} = 1\]
\[ \Rightarrow \frac{d}{2} = 1\]
\[ \Rightarrow d = 2\]
\[\text { Putting the value in (i), we get }: \]
\[a = \frac{3 \times 2}{2}\]
\[ \Rightarrow a = 3\]
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