if 4th term of an arithmetic progression is 3 times the 1st term nd 7th term exceeds twice the third term by 1. find the first term and common difference

a=5 and d=4

but the answer is not clear, how the first and common difference is solve

To find the first term and common difference of an arithmetic progression, we can follow these steps:

Step 1: Let's assume the first term of the arithmetic progression is "a" and the common difference is "d".

Step 2: Since the 4th term is 3 times the 1st term, we can write it as:
a + 3d = 3a ... (Equation 1)

Step 3: Similarly, since the 7th term exceeds twice the third term by 1, we can write it as:
a + 6d = 2(a + 2d) + 1 ... (Equation 2)

Now, we have a system of two equations with two unknowns (a and d). We can solve this system of equations to find the values of a and d.

Let's simplify and solve these equations step by step:

First, let's solve Equation 1 for 'a':
a + 3d = 3a
3d = 2a
a = (3d) / 2

Now substitute this value of 'a' into Equation 2:
(3d) / 2 + 6d = 2((3d) / 2 + 2d) + 1

Simplify and solve for 'd':
(3d + 12d) / 2 = (3d + 4d) + 1
15d / 2 = 7d + 1
15d = 14d + 2
15d - 14d = 2
d = 2

Now substitute the value of 'd' back into Equation 1 to find 'a':
a + 3(2) = 3a
a + 6 = 3a
2a = 6
a = 3

Therefore, the first term is 3, and the common difference is 2.

T4 = 3T1

T7 = 2T3 + 1

Now just plug in the values for Tn and you have

a+3d = 3a
a+6d = 2(a+2d)+1

Now just solve for a and d.