Find an arithmetic progression for which $\begin{cases}S_n-a_1=48\\S_n-a_n=36\\S_n-a_1-a_2-a_{n-1}-a_n=21\end{cases}$

Question

Find an arithmetic progression for which $$\begin{cases}S_n-a_1=48\\S_n-a_n=36\\S_n-a_1-a_2-a_{n-1}-a_n=21\end{cases}$$

I have tried to use the formula $$S_n=\dfrac{a_1+a_n}{2}.n$$ but it seems useless at the end. From the first equation $$a_1=S_n-48$$ and I tried to put it into the third, but it isn't very helpful. Thank you!

Answer 1

If we put the first equation into the third, we have

$$48-a_2-a_{n-1}-a_n=21$$

$$a_2+a_{n-1}+a_n = 27$$

$$3a_1+d+(n-2)d+(n-1)d=27$$

$$3a_1+(2n-2)d=27$$

$$3a_1+2(n-1)d=27\tag{A}$$

If we subtract the first two equations, we have

$$a_n-a_1=12$$

$$(n-1)d=12\tag{B}$$

From $(A)$ and $(B)$, we can solve for $a_1=1$.

Now substituting $a_1$ into the first equation, we can solve for $S_n=49$ and from the second equation$a_n = S_n-36=13$.

$$S_n = \frac{n}{2}(1+a_n)=49$$

$$n=\frac{98}{14}=7.$$

I will leave finding the common difference to you.

Answer 2

Notice the useful simplification $$\color{green}{a_1+a_2+a_{n-1}+a_n=2a_1+2a_n}.$$

So from $(1)-(2)$,

$$a_n-a_1=12$$

and from $\dfrac{(1)+(2)-2\times(3)}3$,

$$a_n+a_1=14.$$

The rest is easy,

$$a_1=1,a_n=13$$ and $$S_n=49,n=7.$$

Answer 3

You can use the property of an A.P. which says that $"$Sum of terms taken symmetrically from both the sides(start and end) is equal. $"$

The A.P. is $a_1,a_2,\cdots , a_{n-1}, a_n$. By above property we have

$a_1+a_n=a_2+a_{n-1}$

Using above property in third equation we get

$S_n-2(a_1+a_n)=21$

Adding first two gives

$2S_n-(a_1+a_n)=84$

Solving above two equation gives

$a_1+a_n=14$

Subtracting second from first equation gives

$a_n-a_1=12$

Therefore $a_1=1$ and $a_n=13$

Also $S_n=49$, Therefore number of terms $\displaystyle n=\frac{2S_n}{a_1+a_n}=7$

$a_n-a_1=(n-1)d\implies 6*d=12\implies d=2$

Therefore you A.P. is $1,3,5,7,9,11,13$

Answer 4

We might also "break down" the given information in this way. If $ \ S_n - a_1 \ = \ 48 \ \ , $ then subtracting $ \ a_n \ $ from the sum instead "costs" $ \ 12 \ $ more, or $$ \ S_n \ - \ a_n \ \ = \ \ 36 \ \ = \ \ S_n \ - \ (a_1 + 12) \ \ = \ \ S_n \ - \ ( \ a_1 + [n-1]·d \ ) \ \ \Rightarrow \ \ (n-1)·d \ = \ 12 \ \ . $$

The third given equation tells us that subtracting the four specified terms from the sum "costs" $ \ 27 \ $ more than subtracting the first term alone, hence $$ S_n \ - \ a_1 \ - \ a_2 \ - \ a_{n-1} \ - \ a_n \ \ = \ \ 21 $$ $$ \Rightarrow \ \ ( \ S_n \ - \ a_1 \ ) \ - \ ( \ a_1 + d \ ) \ - \ ( \ a_1 + [n-2]·d \ ) \ - \ ( \ a_1 + [n-1]·d \ ) \ \ = \ \ 21 $$ $$ \Rightarrow \ \ 48 \ - \ 3· a_1 \ - \ 2 · [n-1]·d \ \ = \ \ 48 \ - \ 3· a_1 \ - \ 2 · 12 \ \ = \ \ 24 \ - \ 3· a_1 \ \ = \ \ 21 \ \ . $$ Therefore $ \ \mathbf{a_1 \ = \ 1} \ \ , \ \ \mathbf{S_n} \ = \ 48 + 1 \ \mathbf{= \ 49} \ \ $ and $ \ a_n \ = \ 49 - 36 \ = \ 13 \ \ . $ We can then use the "first-and-last-term" sum formula to determine that $$ S_n \ \ = \ \ \frac{n}{2}·(a_1 \ + \ a_n) \ \ \Rightarrow \ \ 49 \ \ = \ \ \frac{n}{2}·(1 + 13) \ \ \Rightarrow \ \ \mathbf{n} \ \ = \ \ 2·\frac{49}{14} \ \ \mathbf{= \ \ 7} \ \ $$ and, thence, $ \ \mathbf{d} \ = \ \frac{12}{7 - 1} \ \mathbf{= \ 2} \ \ . $

(Each of the posted answers amounts to different ways of seeing the given conditions in terms of the relations for an arithmetic progression.)