Number of terms in an Arithmetic Sequence

Question

Here is a question -

The sum of the first four terms of an Arithmetic Sequence is 56. The sum of the last four terms is 112. If its first term is 11, then find the number of terms.

Here's what I have done so far -

To find the common difference, $d$ -

$\Rightarrow\;$ Sum = $\displaystyle \frac{n}{2}(2a+(n-1)d)$

$\Rightarrow\; \displaystyle \frac{4}{2}(2(11)+(4-1)d)=56$

$\Rightarrow\; \displaystyle 22+3d=28$

$\Rightarrow\; \displaystyle 3d=6$

$\displaystyle \therefore\ d=2$

Now to find the first term, $a_1$, of the last four terms -

$\Rightarrow\; \displaystyle \frac{4}{2}(2(a_1)+2(4-1))=112$

$\Rightarrow\; \displaystyle 2a_1+6=56$

$\Rightarrow\; \displaystyle 2a_1=50$

$\displaystyle \therefore\ a_1=25$

Then, to find the last term of the sequence -

$\Rightarrow\; \displaystyle a_4=a_1+3d$

$\Rightarrow\; \displaystyle a_4=25+3(2)$

$\displaystyle \therefore\ a_4=31$

Now I don't know what to do after this. Given the first term ($11$) and last term ($31$) of the sequence, how would you find the number of terms, $n$, in this sequence?

Answer 1

Proceeding from what you've already done, write $a_1 = 11$ and $a_n=31$. You have the recurrence $a_r = a_{r-1}+2$ and so $a_n = a_1 + (n-1)d$.

The rest is just evaluating $31 = 11 + 2(n-1) \implies n = 11$

Answer 2

Assume there are an odd number of terms (say $2n+1$) Then, let the middle term be $a$

$(a-nd)+(a-(n-1)d)+(a-(n-2)d)+(a-(n-3)d)=56$

$(a+nd)+(a+(n-1)d)+(a+(n-2)d)+(a-(n-3)d)=112$

(add)

$8a=56+112=168\implies a=21$

Now, Subtract the equations

$2nd+2(n-1)d+2(n-2)d+2(n-3)d=112-56=56$

$8nd-12d=56\implies 2nd-3d=14\implies2nd=3d+14$

Also, $21-nd=11\implies nd=10$

So, $d=2$

$21-nd=11\implies n=5$

Number of terms $=2n+1=11$

EDIT: This method works even if there are an even number of terms. Simply let the number of terms be $2n+1$ where $n$ is a fraction, and $a$ is the average of the middle pair