Given a sequence ($u_1; u_2; u_3;...; u_n$), $n\in \mathbb{N}$, of $n$ terms in geometric series, show that for every natural number $0\le p\le n-1$,

$$u_{1+p} \cdot u_{n-p}=u_1 \cdot u_n$$

I know I'm not supposed to ask for solutions for textbooks exercises, but I'm teaching myself math without ever have had a math class in highschool. I would appreciate not only a solution but an intuitive description.

  • 1
    $\begingroup$ Hint: Use the fact that $u_n = r^n$, try writting your equation using this fact $\endgroup$ – JoseSquare Feb 11 '19 at 18:08
  • $\begingroup$ But isn't it $u_1 \cdot r^n$ ? $\endgroup$ – Daniel Oscar Feb 11 '19 at 18:10
  • $\begingroup$ @DanielOscar You want $u_n = u_0 r^n$---notice that this formula gives an identity in the special case $n = 0$. $\endgroup$ – Travis Willse Feb 11 '19 at 18:14
  • $\begingroup$ Doesn’t matter, look at the answer Dr.Sonnhard post $\endgroup$ – JoseSquare Feb 11 '19 at 18:14

We will use the fact, that $$u_n=u_1q^{n-1}$$ so $$u_{n-p}=u_1q^p$$ and $$u_{n-p}=u_1q^{n-p-1}$$ and we get $$u_{1+p}\cdot u_{n-p}=u_1^2q^{n-1}$$ and the right-hand side: $$u_1\cdot u_n=u_1^2q^{n-1}$$ and this is the same term.


The definition of a geometric sequence is that $u_{k+1} = r*u_{k}$ for a constant $r$.

So $u_{1+p} * u_{n-p} = $

$(u_p*r)*u_{n_p} = u_p*(r*u_{n-p})=$

$u_p*u_{n-p+1} = $

$(u_{p-1}*r)*u_{n-p + 1} = u_{p-1}*(r*u_{n-p + 1}) =$

$u_{p-1}*u_{n-p + 2}= $


$u_{p-k}*u_{n-p + k+1}=$



$u_1*r*u_{n-1} =$



It's worth noting Via induction $u_1$ (the first term) is some constant.

$u_2 = r*u_1$ and $u_3 = r*u_2 = r*(r*u_1) = r^2*u_1$ and $u_4= r*u_3 = r(r^2*u_1)=r^3*u_1$ and inductively so on: $u_{k+1} = r^k*u_1$.

Which in turn means:

$r^k*u_m = r^k*(r^{m-1}*u_1) = r^{k+m-1}*u_1 = u_{k+m}$.


$u_{1+p}*u_{n-p} = r^p*u_1*r^{n-p-1}*u_1 = u_1*r^{n-1}*u_1 = u_1*u_n$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.