Substituting a = (3)/(2)d from equation (1) into 2d - a = 1, we obtain (1)/(2)d = 1, which gives d = 2. Then a = (3)/(2) × 2 = 3.

the fourth term of an ap is 3 times its first term and the Seven term exceed twice the third term by one find AP

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

Step-by-step solution

Step 1: Use general term formula to set up the first condition

Let the first term of the arithmetic progression be $a$ and the common difference be $d$ . The $n$ -th term is given by $a_n = a + (n-1)d$ . The first given condition states that $a_4 = 3a$ .

Step 2: Establish the relation between a and d

Substituting $a_4 = a + 3d$ into $a_4 = 3a$ , we get $a + 3d = 3a$ . Subtracting $a$ from both sides gives $2a = 3d$ , which can be labeled as equation (1).

Step 3: Set up the equation for the second condition

The seventh term is $a_7 = a + 6d$ and the third term is $a_3 = a + 2d$ . According to the problem, the seventh term exceeds twice the third term by $1$ , so $a_7 - 2a_3 = 1$ .

Step 4: Simplify the second equation

Expanding the right-hand side gives $2a + 4d + 1$ . Rearranging terms by subtracting $a$ and $4d$ from both sides yields $2d - a = 1$ .

Step 5: Solve for the common difference d and first term a

Substituting $a = \frac{3}{2}d$ from equation (1) into $2d - a = 1$ , we obtain $\frac{1}{2}d = 1$ , which gives $d = 2$ . Then $a = \frac{3}{2} \times 2 = 3$ .

Step 6: Write down the arithmetic progression

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

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