Half the infinite sum and alternate indices

(a) In a convergent geometric progression, the sum of the first nn terms equals the sum of the remaining terms. Given also that the (n+1)(n+1)th term is 12\frac12, determine the (2n+1)(2n+1)th term. marks

(b) An arithmetic progression has first term aa and common difference dd. The eighth, third and second terms are successive terms of an infinite geometric progression. If the first GP term is 10, find the sum of its even-numbered terms. marks

A sequence is formed whose nnth term is ln⁡∣unun+1∣\ln|u_nu_{n+1}|, where unu_n is the nnth term of the geometric progression. Show that it forms an AP and state the common difference. marks

[2009/NYJC/Prelim/I/9]

Total marks: 11

Solution

(a)

Equal halves

The sum of the first nn terms equals the sum of the remaining terms and the (n+1)(n+1)th term is 12\frac12. Find the (2n+1)(2n+1)th term.

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Hint

  1. Sn=S∞−Sn⇒2Sn=S∞S_n=S_\infty-S_n\Rightarrow2S_n=S_\infty.

Answer · (a)

1/41/4.

Explore the idea

Case n = 1: a = 1 and r = 1/2

Drag the slider to change k and compare the plotted values.

Indexk = 1
Try an example
term uₖ at k = 1
1
sum Sₖ at k = 1
1
Key idea

With rn=12r^n=\frac12 the first nn terms add to exactly half of S∞=2S_\infty=2.

Ideas to try
Check that S1=1S_1=1 equals the sum of all the remaining terms.
(b)

Even-indexed sums and logarithms

Find the sum of the even-indexed GP terms when its first term is 10. Then show that tn=ln⁡∣unun+1∣t_n=\ln|u_nu_{n+1}| forms an AP and state its common difference.

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Hint

  1. Equal ratios: a+2da+7d=a+da+2d\frac{a+2d}{a+7d}=\frac{a+d}{a+2d}.

  2. Even-numbered terms form a GP with ratio r2r^2.

  3. tn−tn−1=ln⁡∣un+1un−1∣t_n-t_{n-1}=\ln\left|\frac{u_{n+1}}{u_{n-1}}\right|.

  4. Write un=10rn−1u_n=10r^{n-1} and simplify the logarithm before subtracting terms.

Answer · (b)

With convergence assumed: even-indexed sum 25/1225/12; logarithmic AP difference −2ln⁡5-2\ln5.

Literal infinite-GP wording also allows the constant GP: even-indexed series diverges and log-AP difference is 0.

Explore the idea

Constant differences after taking logarithms

For the intended ratio 1/5, the log-products form an AP. The vertical axis shows log-products only.

Indexn = 5
Reset the controls
ln(uₙuₙ₊₁) at n = 5
-9.8798
Key idea

The terms shrink geometrically but their log-products fall by the same amount each time: −2ln⁡5≈−3.22-2\ln5\approx-3.22.

Ideas to try
How does tnt_n change as nn increases by 1?
Takeaway

Subsequence first term and ratio both change

Even-indexed GP terms start at arar and use ratio r2r^2; odd-indexed terms start at aa with the same squared ratio.