The fourth term of an A.P is equal to 3 times the first term and the seventh term exceeds twice the third term by 2. The first term and the common difference are:
a+3d = 3a
a+6d = 2(a+2d)+2
now just crank it out.
The first term of ap exceed twice the second term by one . The 10th term exceed twice the fourth term by 3 find the first term
To find the first term and the common difference of the arithmetic progression (A.P.), let's assign variables:
Let the first term be 'a' and the common difference be 'd'.
Given:
The fourth term (a₄) is equal to 3 times the first term (a):
a₄ = 3a ---(1)
The seventh term (a₇) exceeds twice the third term (2a₃) by 2:
a₇ = 2a₃ + 2 ---(2)
To find 'a' and 'd', we need to solve equations (1) and (2).
Let's start by finding the fourth term (a₄) using equation (1).
In an arithmetic progression, the nth term is given by: aₙ = a + (n-1)d
So, substituting n = 4 into the formula, we get:
a₄ = a + (4 - 1)d
a₄ = a + 3d
Equation (1): a₄ = 3a
Therefore, we can write: a + 3d = 3a
Rearranging the equation, we get: 2a = 3d
Divide both sides by 3: a = (3/2)d ---(3)
Now let's find the seventh term (a₇) using equation (2).
Substituting n = 7 into the formula for the nth term, we have:
a₇ = a + (7 - 1)d
a₇ = a + 6d
Equation (2): a₇ = 2a₃ + 2
Replacing a₃ with a + 2d, we get: a + 6d = 2(a + 2d) + 2
Expanding and simplifying, we have: a + 6d = 2a + 4d + 2
Rearranging the equation, we get: 4d = a + 2
Substituting equation (3) in equation 4d = a + 2, we have:
4d = (3/2)d + 2
To solve for 'd':
Multiply both sides by 2 to clear the fraction:
8d = 3d + 4
8d - 3d = 4
5d = 4
Divide both sides by 5: d = 4/5
Now, substitute the value of 'd' into equation (3):
a = (3/2)d
a = (3/2)(4/5)
a = 12/10
a = 6/5
Therefore, the first term (a) is 6/5 and the common difference (d) is 4/5 in the arithmetic progression.