MHB Sum of pqth Term in Arithmetic Progression

AI Thread Summary
The discussion focuses on finding the sum of the pqth term in an arithmetic progression where the pth term is 1/q and the qth term is 1/p. By establishing equations for the first term (a1) and the common difference (d), both are found to equal 1/pq. Using the sum formula for an arithmetic progression, S_n = n/2(2a1 + (n-1)d), with n set to pq, the sum simplifies to pq. Ultimately, the sum of the pqth term is confirmed to be pq.
Doffy
Messages
12
Reaction score
0
The pth term of an airthmetic progression is 1/q and qth term is 1/p. What is the sum of pqth term?
 
Mathematics news on Phys.org
I would begin by stating:

$$a_p=a_1+(p-1)d=\frac{1}{q}$$

$$a_q=a_1+(q-1)d=\frac{1}{p}$$

Now, solve both for $a_1$...then equate the results and solve that for $d$...what do you find?
 
According to this, we find that a1=d=1/pq.
However, I am still confused about the sum of pqth term.
 
Doffy said:
According to this, we find that a1=d=1/pq.
However, I am still confused about the sum of pqth term.

Yes, that's correct. So now apply the formula:

$$S_n=\frac{n}{2}\left(2a_1+(n-1)d\right)$$

where:

$$n=pq,\,a_1=d=\frac{1}{pq}$$
 
What we then get is:

$$S_{pq}=\frac{pq}{2}\left(\frac{2}{pq}+\frac{pq-1}{pq}\right)=1+pq-1=pq$$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Back
Top