First 4 terms of an arithmetic series, given $S_{10}$ and $S_{14}$

Question

I've been stuck on this one for a solid half hour, and it proved too niche to google.

Given $S_{9}=144$ and $S_{14}=329$ of an arithmetic series, and I'm required to find $t_{1},\cdots,t_{4}$.

I've tried using $S_{n}={n\over2}[2a+(n-1)d]$ to rearrange for $a$ and $d$, as well as $d={S_{14}-S_{9}\over5}$, but I've gotten nowhere.

Edit: the answer in the back of the package is $4, 7, 10, 13$

Answer 1

It should be $\displaystyle d=\frac{T_{14}-T_5}{5}$ instead of $\displaystyle d=\frac{S_{14}-S_5}{5}$.

We have

$$ 144=S_{9}=\frac{9}{2}(2a+8d)=9a+36d$$

and

$$ 329=S_{14}=\frac{14}{2}(2a+13d)=14a+91d$$

So

\begin{align} 14\times 144-9\times 329 &=(14\times 36-9\times91)d\\ d&=3 \end{align}

Therefore, $\displaystyle a=\frac{144-36\times 3}{9}=4$.

$T_1=4$, $T_2=7$, $T_3=10$ and $T_4=13$.

Answer 2

you will get $$S_9=\frac{9}{2}\left(2a_1+8d\right)=144$$ $$S_{14}=\frac{14}{2}\left(2a_1+13d\right)=329$$ can you solve this? after simplification we obtain $$2a_1+8d=32$$ $$2a_1+13d=47$$ by multipliying the first equation with $-1$ we have $$-2a_1-8d=-32$$ $$2a_1+13d=47$$ adding both we get $$5d=15$$ can you finish now?

Answer 3

$S_9=144$ means that the average of the first $9$ terms is $144/9=16$. So $t_5=16$.

Similarly, $S_{14}=329$ means that the average of terms $t_{10},\ldots,t_{14}$ is $(329-144)/5=37$. So $t_{12}=37$.

Now you have two terms of the sequence, so you can easily compute $d$ and $a$.

Answer 4

This method is similar to @TonyK's but with a slight difference, using a virtual middle term for an AP with an even number of terms. The same concept still works given the linearity of the AP.

$$T_5=\frac {S_9}9=16\\ T_{7.5}=\frac {S_{14}}{14}=23.5\\ T_{7.5}-T_5=2.5d=7.5\\ d=3\\ a=4$$ Hence $$T_1, T_2, T_3, T_4=4,7,10,13$$