the fourth term of an arithmetic progression is equal to 3 times the first term and the seventh term exceeds twice the third term by 1 find the first term and the common differrence
a+3d = 3a
a+6d = 2(a+2d)+1
a=3
d=2
chec:
3+6 = 3*3
3+12 = 2(7)+1
To find the first term and the common difference of the arithmetic progression, let's break down the given information step by step.
Let's assume that the first term of the arithmetic progression is represented by 'a', and the common difference is represented by 'd'. Therefore, the terms of the arithmetic progression can be expressed as:
First term: a
Second term: a + d
Third term: a + 2d
Fourth term: a + 3d
Fifth term: a + 4d
Sixth term: a + 5d
Seventh term: a + 6d
Given information:
The fourth term (a + 3d) is equal to 3 times the first term (3a):
a + 3d = 3a ----------(1)
The seventh term (a + 6d) exceeds twice the third term (2(a + 2d)) by 1:
a + 6d = 2(a + 2d) + 1 ----------(2)
Now, let's solve these two equations to find the values of 'a' and 'd'.
From equation (1), we can simplify it as:
3d = 2a
Rearranging the terms:
2a - 3d = 0 ----------(3)
From equation (2), let's expand the equation:
a + 6d = 2a + 4d + 1
Simplifying the equation, we get:
2d - a = 1 ----------(4)
Now, we have two equations, (3) and (4), which we can solve simultaneously to find the values of 'a' and 'd'.
To solve these equations, we can use the method of substitution or elimination. Here, we'll use the method of substitution:
From equation (3), we can express 'a' in terms of 'd':
a = (3/2)d ----------(5)
Substitute this value of 'a' in equation (4):
2d - (3/2)d = 1
Simplifying and rearranging:
(4/2)d - (3/2)d = 1
(1/2)d = 1
Multiplying both sides by 2:
d = 2
We have found the value of the common difference 'd' as 2.
Now, substitute the value of 'd' into equation (5) to find 'a':
a = (3/2)(2)
a = 3
Hence, the first term (a) of the arithmetic progression is 3, and the common difference (d) is 2.