The sum of first term and seventh term of an arithmetic progression is 16.if the twentieth term is 56,find the term whose value first exceeds 1000.

Write the equation first
The sum of the first and seventh is 16
a₁ + a₇ = 16
a + (a + (n-1)d) = 16
a + (a + (7-1)d) = 16
2a + 6d = 16 (simplify)
a + 3d = 8
This is equation (1)

The twentieth term is 56
a₂₀ = 56
a + (n-1)d = 56
a + (20-1)d = 56
a + 19d = 56
This is equation (2)

Eliminate a in equation (1) and (2) to find d
a + 19d = 56
a +   3d =  8
- (substract)
       16d = 48
           d = 3

After finding d, we subtitute the number of d to one of the equation
a + 3d = 8
a + 3(3) = 8
a + 9 = 8
a = -1

Now, we can find the general formula for "an" (n term)
an = a + (n-1)d
an = -1 + (n-1)3
an = -1 + 3n - 3
an = 3n - 4
This is the general formula of this arithmetic progression

We could find the first term that exceeds 1,000 by making inequality
an > 1,000
3n - 4 > 1,000
3n > 1,004
n > 334.67
the first term that exceeds 1,000 is the 335th term

Aso sorry if i’m wrong