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Central symmetry in a GP

Authors
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    Name
    Vu Hung
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Problem Statement

In a GP, define Pn=T1T2⋯TnP_n=T_1T_2\cdots T_n.

  • Show Pn=anrn(n−1)2P_n=a^n r^{\frac{n(n-1)}2}.
  • Show TkTn−k+1=T1TnT_kT_{n-k+1}=T_1T_n.
  • If n=11n=11 and middle term T6=5T_6=5, find P11P_{11}.

Hints

Use Tk=ark−1T_k=ar^{k-1} and pair symmetric terms.

Solutions

Pn=a(ar)⋯(arn−1)=anr0+1+⋯+(n−1)=anrn(n−1)2.P_n=a(ar)\cdots(ar^{n-1})=a^n r^{0+1+\cdots+(n-1)}=a^n r^{\frac{n(n-1)}2}.

Also

TkTn−k+1=ark−1⋅arn−k=a2rn−1=T1Tn.T_kT_{n-k+1}=ar^{k-1}\cdot ar^{n-k}=a^2r^{n-1}=T_1T_n.

For odd n=11n=11, the product is middle-term power:

P11=T611=511.P_{11}=T_6^{11}=5^{11}.

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