The second term of an arithmetic sequence is $13$ and $5^{th}$ term is $31$. What is the $17^{th}$ term of the sequence?

$4.$ Evaluate the following:

$a.$ The second term of an arithmetic sequence is $13$ and $5^{th}$ term is $31$. What is the $17^{th}$ term of the sequence?

I am trying to do this question (see above). Am I supposed to use simultaneous equations to solve for $a$ and $d$? And then from there sub the values for $a$ and $d$ into the arithmetic sequence rule to then solve for X_{17}? However, when I did this it did not work:

Using arithmetic rule, $X_n=a+d(n-1)$, I attained the following: $$X_2 = 13 \longrightarrow a+d = 13 \\ X_5 = 31 \longrightarrow 4(a+d) = 31$$ I then isolated both $a$ and $d$ in both equations and attained the following when subbing one equation into the other: $$a = 13-(7.75-a) \\ d = 7.75-(13-d)$$ However, when I solve for both $a$ and $d$, it seems impossible as they just subtract leaving me with nothing.

• "I attained the following"... X5 is totally wrong. Commented Jul 12, 2016 at 8:32

According to question:

$$(a+d)=13 \dots (1)$$

$$(a+4d)=31 \dots (2)$$

On solving equation $(1)$ and $(2)$, we get

First term $a=7$, and common difference $d=6$.

So, $17^{th}$ term is $=a+16d=7+16\times 6=103$(Answer).

Sequence is $7, 13, 19, 25, 31, 37, \dots$

There's an error in your second equation: it should be

$$a+4d=31$$

which, combined with the first, yields $a=7, d=6$.

Substitute in $X_{17} = a+16d$ to get the answer.

Your calculation is wrong... Although it started well. Lets keep $X_n=a+dn$ (using $n-1$ instead of $n$ does not have much interest here).

Then you stated that $X_2=13$ and $X_5=31$.

So we can write :

\begin{align} X_2=13=a+2d &~~~~~~(L_1)\\ X_5=31=a+5d &~~~~~~(L_2) \end{align}

We just replaced $n$ with $2$ and $5$ in our starting expression.

Now, substracting $L_2-L_1$, we get $18=3d$, so we deduce that $d=6$.

And in $L_1$, we now have $13=a+2\times6$ so $a=1$.

Finally, we can write $X_n$ for all $n$ :

$$X_n=1+6n$$

And $X_{17}$ comes easily :

$$X_{17}=1+6\times17=103$$