Eliminate two progression parameters
An arithmetic series has first term and common difference , where are non-zero. A convergent geometric series has first term and common ratio , where is positive and is non-zero. The fourth and ninth terms of the arithmetic series equal the sixth and ninth terms of the geometric series respectively, and the eleventh term of the arithmetic series is less than the fourteenth term of the geometric series.
(i) Show satisfies and solve, giving your answer correct to 4 decimal places. marks
(ii) Using this value of , deduce that for any positive integer , the sum of the terms of the geometric series after, but not including, term is less than . marks
[2024/NJC/Prelim/II/2]
Solution
Eliminating parameters
Show and solve it, giving to 4 d.p.
Hint
, , .
Subtract to find two ways.
A convergent series needs .
Answer · (i)
; convergent root (4 d.p.).
Three equations
4th, 9th, 11th AP terms and 6th, 9th, 14th GP terms.
Two expressions for
Subtract pairs.
Equate
Divide by .
Solve
Real roots 1.1371 and 0.3435; convergence needs .
Explore the idea
Bounding the tail
Deduce that for every positive integer the sum of the terms after term is less than .
Hint
Tail .
Its largest value is at .
Here the bound has plenty of room.
Answer · (ii)
Tail .
; use the unrounded root.
The tail
Terms form a GP with first term . Alternatively subtract from to obtain the same tail.
Largest at
The admissible root satisfies and , so the tail decreases with . The largest allowed tail occurs at ; evaluate with the stored root.
Compare
Explore the idea
A bound for every n needs a maximum
Checking a few tails is not enough. Prove the tail decreases, identify its largest allowed value, then compare that value with the required bound.