An arithmetic sequence has first term a and common difference d - AQA - A-Level Maths Pure - Question 9 - 2018 - Paper 1

To find the sixth term of the sequence, we use the expression for the nth term: $t_n = a + (n - 1)d$

For the sixth term, where $n = 6$: $t_6 = a + 5d$

We know from the problem statement that $t_6 = 25$: $a + 5d = 25$

Rearranging this gives: $a = 25 - 5d$

Now, substituting this expression for $a$ into the quadratic equation we derived in part (a): $4(25 - 5d) + 70d = 4(25 - 5d)^2 + 20(25 - 5d)d + 25d^2$

Expanding and simplifying gives:

1. Substitute into $4a + 70d = 4a^2 + 20ad + 25d^2$.
2. Minimize the obtained expression for $d$.

Eventually, setting $d = 5$ will yield: $a = -55$

Verifying with the quadratic equation shows that this is the smallest achievable value for $a$. Thus, the solution yields: \smallest possible value of a = -55.