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.
$a_{4} = 3 a \Rightarrow a + (4 - 1) d = 3 a \Rightarrow a + 3 d = 3 a \Rightarrow 3 d = 2 a \Rightarrow a = \frac{3 d}{2} \dots \dots (i)$
And, $a_{7} - 2 a_{3} = 1$
$\Rightarrow a + (7 - 1) d - 2 [a + (3 - 1) d] = 1 \Rightarrow a + 6 d - 2 (a + 2 d) = 1 \Rightarrow a + 6 d - 2 a - 4 d = 1 \Rightarrow - a + 2 d = 1 \Rightarrow - \frac{3 d}{2} + 2 d = 1 [From (i)] \Rightarrow \frac{- 3 d + 4 d}{2} = 1 \Rightarrow \frac{d}{2} = 1 \Rightarrow d = 2$
Putting the value in (i), we get
$a = \frac{3 \times 2}{2} \Rightarrow a = 3$

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