I was wondering about this when it hit me, can a sequence ever be both arithmetic and geometric? I was thinking maybe a sequence like 0, 0, 0, 0... or 1, 1, 1, 1... where it's constant but I don't know thoroughly if there are any restrictions on arithmetic and geometric sequences that prohibit situations like these. Could this be possible? My second question is: An arithmetic sequence with general term A(n), n>=1 has the sum of the first two terms 5 and the sum of third and fourth terms 17. Find [tex]\sum_{i=1}^{10}{Ai} [/tex] I don't really have an idea of how to approach this. We know that: [tex]A1 + A2 = 5[/tex] [tex]A3 + A4 = 17[/tex] But in the arithmetic sequence, how would you find [tex]a[/tex] or [tex]d[/tex]?
But how do I find the sequence though? It says that arithmetic progression is just another fancy term for an arithmetic sequence. GIven the information that I have, how would I found the first term and the constant? I know how to obtain the overall answer from a certain point in the problem but I don't know how to start it. As long as I can find the first term of the sequence and the constant, it's all I need. But I don't know how.
If you are asking whether a sequence can be of the form [tex]a_{n+1} = r a_n + d[/tex] then, of course, the answer is yes. Luckily, this can be transformed into a geometric sequence with the tranformation [itex]a_n = b_n + c[/itex] with a suitable choice for c so that [itex]b_{n+1} = r b_n[/itex]. HINT: For your second question, recognize that [itex]a_n = a_1 + (n-1)d[/itex].
So would I initially start off with that question like this? [tex] a_2 = (5 - a+2) + (2-1)d [/tex] Would it be possible to create 2 system of equations then using elimation to find [tex]a_1[/tex] and the the constant? Ack, I still don't understand. I can't thinkof any way that I can use the arithmetic sequence formula to solve for a1 or d.
Start simply. An arithmetic sequence (or series) is always of the form a_{n}= a_{1}+ (n-1)d. Here, n-1 just counts the number of times the "common difference" has been added to the first term. You are told "the sum of the first two terms 5". Okay, the first term is, of course, a_{1} and the first term is just that plus the difference: a_{1}+ d. The sum of the first two terms is a_{1}+ (a_{1}+ d)= 2a_{1}+ d= 5. You are told "the sum of third and fourth terms 17". The third term is a_{1}+ 2d and the fourth term is a_{1}+ 3d so the sum of the third and fourth terms is (a_{1}+ 2d)+ (a_{1}+ 3d)= 2a_{1}+ 5d= 17. You can solve the two equations 2a_{1}+ d= 5 and 2a_{1}+5d= 17 for a_{1} and d (I recommend subtracting one equation from the other to find d). Then it is easy to find the sum.
I hate myself gravely for not being able to see simple things like that, or what appears to be simple things. It looks so hard from every perspective, but when you see the solution, it seems so simple. I just have 1 small question HallsofIvy, when you find the summation of the first 10 terms after finding A1 and you know d, is it 145? I don't have an answer for this question since it was assigned spontaneously by my professor. And thank you HallsofIvy, words cannot express how grateful I am of your help.
[itex]a_n = a_0 + n d[/itex] so [itex]\sum_{k=1}^{10} a_k = 10a_0 + d\sum_{k=1}^{10} k[/itex]. You should be able to figure out the sum.
(Finite) Arithmetic sequences have the nice property that the average of ALL terms in the sequence is the same as the average of the first and last number. And, of course, adding all n numbers in the sequence is the sequence is the same as multiplying that average by n. If a_{n}= a_{1}+ (n-1)d, then the average of the first and last terms is [tex]\frac{a_1+ (a_1+ (n-1)d)}{2}= \frac{2a_1+ (n-1)d}{2}= a_1+ \frac{n-1}{2}d[/tex] The sum of the first n terms is then [tex]n\left(a_1+ \frac{n-1}{2}d\right)= na_1+ \frac{n(n-1)}{2}[/tex] That is the same as your formula, assuming (which is correct in this case) that a_{1}= 1. Since you gave the correct answer earlier, I am going to go ahead and give the complete solution. In your particular example, the sum of the first two terms, a_{1} and a_{2}= a_{1}+ d, is 2a_{1}+ d= 5 and the sum of the third and fourth terms, a_{3}= a_{1}+ 2d and a_{4}= a_{0}+ 3d, is 2a_{1}+ 5d= 17. Subtracting the first of those equations from the first, as I suggested, eliminates a_{1} and gives 4d= 12 so d= 3. Putting that into the first equation gives 2a_{1}+ 3= 5 so a_{1}= 1. That is, a_{n}= 1+ 3(n-1). The first 10 terms of the sequence are 1, 4, 7, 10, 13, 16, 19, 22, 25, 28. It's not too difficult to add those up and get 145. However, it is easier to just note that the average of 1 and 28, 29/2, is the average of the entire sequence so its sum is 10(20/2)= 5(29)= 145. By the way, your original question asked whether it was possible for a sequence to be both a geometric and arithmetic sequence and Tide suggested a_{n+1}= ra_{n}+ d which, while a combination of arithmetic and geometric sequences, is itself generally neither. In order to be an arithmetic sequence, we would have to have a_{n}= a_{1}+ (n-1)d and a_{n}= a_{1}r^{n-1}. Dividing one equation by the other gives [tex]\frac{a_1+ (n-1)d}{a_1r^{n-1}}= 1[/tex] or [tex]a_1+ (n-1)d= a_1r^{n-1}[/tex] Taking n= 2 gives [tex]a_1+ d= a_1r[/tex] so that [tex]d= (1-r)a_1[/tex] and taking n= 3 gives [tex]a_1+ 2d= a_1r^2[/tex] so that [tex]d= a_1 \frac{1- r^2}{2}= (1-r)a_1\frac{1+r}{2}[/tex]. Those can only be the same if 1) a_{1}= 0 in which case d= 0 and r can be anything: the sequence 0, 0, 0,... 2) r= 1, in which case d= 0 and a_{1} can be anything: the sequence a, a, ,a ... 3) r= 2, in which case d= 0 and a_{1} can be anything: again a, a, a,... That is the constant sequences are the only sequences that are both geometric (with r= 1) and arithmetic (with d= 0).
To address the first question, which I think tide misinterpreted... You can not have a sequence of more than 2 terms (other than the trivial case of the constant sequence) that is both an AP and a GP. Proof : Consider 3 consecutive terms of a GP. Let them be a, ar and ar^{2}. If these 3 terms also formed an AP, the difference between the first 2 terms would equal the difference between the next two terms. In other words we would have : [tex] ar -a = ar^2 - ar [/tex] [tex]\implies a(r-1) = ar(r-1) [/tex] [tex]\implies a=0~or~r=1 [/tex] Both of the above results correspond to trivial sequences. So, no other sequence of 3 or more terms can be both a GP and an AP.
Gokul, I was offering the OP the opportunity to define what he meant by something being "both AP and GP!" :)