Question

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.

Answer 1

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