If the 7th term of an AP is twice the third term and the sum of the first four terms is 42
I assume you're looking for a and d, eh?
a+6d = 2(a+2d)
(4/2)(2a+3d) = 42
Now just solve for a and d.
To solve this problem, we need to use the formula for the nth term of an arithmetic progression (AP), which is given by:
an = a1 + (n-1)d
where an represents the nth term, a1 represents the first term, n represents the term number, and d represents the common difference between consecutive terms.
Given that the 7th term (a7) is twice the third term (2a3) and the sum of the first four terms is 42, we can set up the equations as follows:
a7 = 2a3 --> a1 + 6d = 2(a1 + 2d)
4a1 + 6d = 42
Now, we have a system of two equations with two variables. We can solve this system to find the values of the first term (a1) and the common difference (d).
To solve the system, we can use the method of substitution or elimination.
Let's use the method of substitution:
From the equation 4a1 + 6d = 42, we can solve for a1:
4a1 = 42 - 6d
a1 = (42 - 6d)/4
Now, substitute the value of a1 in the first equation:
a1 + 6d = 2(a1 + 2d)
(42 - 6d)/4 + 6d = 2((42 - 6d)/4 + 2d)
Simplify the equation:
(42 - 6d)/4 + 6d = (42 - 6d)/2 + 4d
Now, cross-multiply to eliminate the fractions:
2(42 - 6d) + 8d = 4(42 - 6d) + 8d
Expand and simplify:
84 - 12d + 8d = 168 - 24d + 8d
Combine like terms:
84 - 4d = 168 - 16d
Move similar variables to one side:
16d - 4d = 168 - 84
12d = 84
Divide by 12:
d = 7
Now, substitute the value of d into either of the original equations:
4a1 + 6(7) = 42
Simplify the equation:
4a1 + 42 = 42
Subtract 42 from both sides:
4a1 = 0
Divide by 4:
a1 = 0
Therefore, the first term (a1) is 0 and the common difference (d) is 7.