Link an AP with a convergent GP

An arithmetic series has first term aa and common difference dd, where a>0a>0 and dd is non-zero. The first, eighth and thirteenth terms of the arithmetic series are equal to the first three terms of a geometric series respectively.

(a) Show that a=−492da=-\frac{49}2d. marks

(b) Show that the geometric series is convergent and find its sum to infinity, SS, in terms of aa. marks

(c) Given that the sum of the first nn terms of the arithmetic series is 25 times the third term of the geometric series, find nn. marks

(d) Find the least kk such that the sum of the first kk terms of the geometric series is more than 98% of SS. marks

[2022/YIJC/Promo/8]

Total marks: 11

Solution

(a)

Equal ratios

The 1st, 8th and 13th terms of an AP are the first three terms of a GP. Show a=−492da=-\frac{49}2d.

Hint

  1. The GP terms are aa, a+7da+7d, a+12da+12d.

  2. Equal ratios: a+7da=a+12da+7d\frac{a+7d}a=\frac{a+12d}{a+7d}.

Answer · (a)

a=−49d/2a=-49d/2.

(b)

Convergence

Show the GP converges and find SS in terms of aa.

Explore this idea

Hint

  1. r=a+7dar=\frac{a+7d}a with d=−2a49d=-\frac{2a}{49}.

Answer · (b)

r=5/7r=5/7, S=7a/2S=7a/2.

Explore the idea

A convergent GP with a = 49 (ratio 5/7)

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

Indexn = 6
Reset the controls
term uₙ at n = 6
9.1108
sum Sₙ at n = 6
148.723
Key idea

The running total tends to 72a\frac72a.

Ideas to try
Slide nn until the sum settles near S=72(49)=171.5S=\frac72(49)=171.5.
(c)

AP sum equals a GP term

The sum of the first nn terms of the AP is 25 times the third term of the GP. Find nn.

Explore this idea

Hint

  1. Third GP term =ar2=2549a=ar^2=\frac{25}{49}a.

  2. AP sum =n2[2a+(n−1)d]=\frac n2[2a+(n-1)d].

Answer · (c)

n=25n=25.

Explore the idea

The AP sum peaks exactly at 625

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

Indexn = 25
Try an example
AP sum Sₙ at n = 25
625
Key idea

The sum of the AP is n(50−n)n(50-n) (with a=49a=49): a parabola whose maximum value 625 is at n=25n=25.

Ideas to try
Why can the AP sum only equal 625 once?
(d)

Fraction of the sum to infinity

Find the least kk such that SkS_k exceeds 98% of SS.

Explore this idea

Hint

  1. Sk>0.98SS_k>0.98S means 1−(57)k>0.981-\left(\frac57\right)^k>0.98.

Answer · (d)

k=12k=12.

Explore the idea

The GP sum against 98% of S

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

Indexn = 12
Try an example
sum Sₖ at n = 12
168.475
Key idea

The sum first exceeds 0.98×171.5=168.070.98\times171.5=168.07 at k=12k=12.

Ideas to try
Compare k=11k=11 and k=12k=12 with the 98% guide.
Takeaway

Keep AP and GP symbols separate

The AP difference is −2a/49-2a/49, while the GP ratio is 5/75/7. Writing each selected term before equating ratios prevents mixing these two roles.