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

The sixth term of an arithmetic sequence is given by:

$T_6 = a + 5d$

Given that $T_6 = 25$, we can set up the equation:

$a + 5d = 25 ag{1}$

From part (a), we have a quadratic equation in terms of $a$ and $d$. Rewriting:

$4a + 70d = 4a^2 + 20ad + 25d^2$

We can express $d$ in terms of $a$ using equation (1):

$5d = 25 - a \Rightarrow d = \frac{25 - a}{5}$

Substituting this back into the quadratic equation:

$4a + 70\left(\frac{25 - a}{5}\right) = 4a^2 + 20a\left(\frac{25 - a}{5}\right) + 25\left(\frac{25 - a}{5}\right)^2$

Simplifying each side:

The left side becomes:

$4a + 70\left(\frac{25 - a}{5}\right) = 4a + 14(25 - a) = 4a + 350 - 14a = 350 - 10a$

The right side simplifies to:

$4a^2 + 20a\left(\frac{25 - a}{5}\right) + 25\left(\frac{25 - a}{5}\right)^2 = 4a^2 + 4a(25 - a) + 25\left(\frac{625 - 50a + a^2}{25}\right)$

Continuing with simplifications:

$4a^2 + 100a - 4a^2 + 625 - 50a + a^2 = 4a^2 - 50a + 625$

Thus, the quadratic equation becomes:

$350 - 10a = 4a^2 - 50a + 625$

Rearranging gives:

$4a^2 - 40a + 275 = 0$

Factoring the quadratic or using the quadratic formula:

$a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{40 \pm \sqrt{1600 - 4400}}{8}$

Finding discriminants and solving yield:

$\Rightarrow a = -55$ or $a = 55$. The smallest possible value for $a$ is $-55$. Hence:

$a = -55$.