Solve the arithmetic progression given the sum of first 4 elements, and the sum of the first and 5th element

Question

I'm trying to solve an arithmetic progression (find the first few elements: a1, a2, a3, ...).

I'm given S4 (sum of the first 4 elements of a sequence) = 14 and the sum of the first a1 and 2 * 5th element, 2*a5 = 0.

Simply put: S4 = 14, a1 + 2a5=0

I tried using the formula (plugging in the values) for the sum of the first n-terms of an arithmetic sequence, but I can't solve it since I don't have neither a1 nor d.

What's the correct way of solving this?

Answer 1

$$S_4 = 14 \implies 2a + 3d = 7 \quad\quad \text{(1.) } $$

$$a+ 2a_5 = 0 \implies a + 2a + 8d = 0\implies a = -\frac{8d}{3} \quad\quad \text{(2.) }$$

Substituting $\text{(2.) }$ in $\text{(1.) }$

$$-\frac{16d}{3} + 3d = 7 \implies -\frac {7d}3 = 7 \implies d =-3 \quad\quad\text{(3.) }$$

From $\text{(2.) }$ and $\text{(3.) }$ $$a = -\frac83 . -3 \implies a=8$$

Hence the A.P is $ \,\, 8,5,2,-1,-4 ....$

Answer 2

Any arithmetic progression is defined by its first term and common difference, so let the $n^{th}$ term and common difference of your desired arithmetic progression be $a_n$ and $d$ respectively. Since we have two unknowns, we need two equations to solve for them.

First, note that $S_n = \frac n2[2a_1 + (n - 1)d]$.

$\because S_4 = 14$,

$\therefore \frac 42[2a_1 + (4-1)d] = 14$

$\Rightarrow 2a_1 + 3d = 7 \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\text{(1)}$

Next, note also that $a_n = a_1 + (n-1)d$

$\Rightarrow a_5 = a_1 + 4d$

$\because a_1 + 2a_5 = 0$,

$\therefore a_1 + 2(a_1 + 4d) = 0$

$\Rightarrow 3a_1 + 8d = 0 \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\text{(2)}$

Given $\text{(1)}$ and $\text{(2)}$, can you solve for $a_1$ and $d$?