The fourth term of an A.P is equal to $3$ times the first term, and the seventh term exceeds twice the $3^{r d}$ term by $1$ . Find the first term and the common difference.

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Solution

Fourth term $= 3 \times$ first term

$\Rightarrow a + 3 d = 3 a$

$\Rightarrow 3 a - a = 3 d$

$\Rightarrow 2 a = 3 d$ or $2 a - 3 d = 0$ ......... $(1)$

Seventh term $- 2 \times$ third term $= 1$

$\Rightarrow a + 6 d - 2 (a + 2 d) = 1$

$\Rightarrow a + 6 d - 2 a - 4 d = 1$

$\Rightarrow - a + 2 d = 1$ ........ $(2)$

$2 \times (2) + 1 \times (1)$

$\Rightarrow - 2 a + 4 d + 2 a - 3 d = 2 + 0 = 2$

$\Rightarrow d = 2$

From $(1)$ we have

$2 a - 3 d = 0$ or $a = \frac{3 d}{2} = \frac{3 \times 2}{2} = 3$ where $d = 2$

$∴$ first term $= 3$ and common difference $= 2$

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