From equation (1), we have a = (3)/(2)d. Substituting this into equation (2) gives (3)/(2)d - 2d = -1, which simplifies to -(1)/(2)d = -1, so d = 2. Substituting d = 2 back gives a = 3.

the fourth term of an ap is 3 times its first term and the 7th term exceed twice 3 3rd term by 1 find the AP

Answer: The AP is $3, 5, 7, 9, \dots$

Step-by-step solution

Step 1: Define the general term of an AP

Let $a$ be the first term and $d$ be the common difference of the arithmetic progression. The formula for the $n^{\text{th}}$ term is $a_n = a + (n-1)d$ .

Step 2: Formulate the first condition

The fourth term is $a_4 = a + 3d$ . According to the given condition, $a_4 = 3a$ , which gives $a + 3d = 3a$ . Rearranging the terms gives $2a - 3d = 0$ , or $2a = 3d$ .

Step 3: Formulate the second condition

The $7^{\text{th}}$ term is $a_7 = a + 6d$ and the $3^{\text{rd}}$ term is $a_3 = a + 2d$ . Given that $a_7$ exceeds $2a_3$ by $1$ , we write $a + 6d = 2(a + 2d) + 1$ . Expanding and simplifying yields $a + 6d = 2a + 4d + 1$ , which simplifies to $a - 2d = -1$ .

Step 4: Solve the system of linear equations

From equation (1), we have $a = \frac{3}{2}d$ . Substituting this into equation (2) gives $\frac{3}{2}d - 2d = -1$ , which simplifies to $-\frac{1}{2}d = -1$ , so $d = 2$ . Substituting $d = 2$ back gives $a = 3$ .

Step 5: Write down the AP

With the first term $a = 3$ and common difference $d = 2$ , the successive terms are obtained by repeatedly adding $2$ . Thus, the required AP is $3, 5, 7, 9, \dots$ .

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