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Difficulty: Very hardArithmetic and Geometric Progressions (AP and GP)

Evaluate the numerical value of each of the following sequence and series expressions, and arrange the items in ascending order (from smallest to largest value):

  1. 1The sum to infinity of the geometric progression 18+6+2+…18 + 6 + 2 + \dots
  2. 2The 4th term of a geometric progression whose 2nd term is 1212 and 5th term is 9696
  3. 3The 15th term of an arithmetic progression whose 3rd term is 1111 and 8th term is 3131
  4. 4The sum of the first 55 terms of an arithmetic progression with first term 44 and common difference 55

Answer

The correct ascending order is: the sum to infinity of 18+6+2+…18 + 6 + 2 + \dots (2727), followed by the 4th term of the geometric progression (4848), then the 15th term of the arithmetic progression (5959), and finally the sum of the first 5 terms of the arithmetic progression (7070).
Evaluating each expression gives values of 2727, 4848, 5959, and 7070. Arranging these from smallest to largest places the sum to infinity (2727) first, the GP term (4848) second, the 15th AP term (5959) third, and the AP sum (7070) last.

Step-by-Step Solution

1
Calculate the sum to infinity of the given geometric progression
For 18+6+2+…18 + 6 + 2 + \dots, first term a=18a = 18 and common ratio r=618=13r = \frac{6}{18} = \frac{1}{3}. Using S∞=a1−rS_\infty = \frac{a}{1-r}, we get S∞=181−1/3=182/3=27S_\infty = \frac{18}{1 - 1/3} = \frac{18}{2/3} = 27.
The sum to infinity formula for a convergent geometric series is S∞=a1−rS_\infty = \frac{a}{1-r} where ∣r∣<1|r| < 1.
2
Calculate the 4th term of the geometric progression
Given T2=ar=12T_2 = ar = 12 and T5=ar4=96T_5 = ar^4 = 96. Dividing T5T_5 by T2T_2 yields r3=9612=8  ⟹  r=2r^3 = \frac{96}{12} = 8 \implies r = 2. Thus a=122=6a = \frac{12}{2} = 6. The 4th term is T4=ar3=6×23=48T_4 = ar^3 = 6 \times 2^3 = 48.
Using the nn-th term formula of a GP, Tn=arn−1T_n = ar^{n-1}, to solve simultaneous equations for aa and rr.
3
Calculate the 15th term of the arithmetic progression
Given T3=a+2d=11T_3 = a + 2d = 11 and T8=a+7d=31T_8 = a + 7d = 31. Subtracting the equations gives 5d=20  ⟹  d=45d = 20 \implies d = 4. Substituting back yields a=11−2(4)=3a = 11 - 2(4) = 3. The 15th term is T15=a+14d=3+14(4)=59T_{15} = a + 14d = 3 + 14(4) = 59.
Using the nn-th term formula of an AP, Tn=a+(n−1)dT_n = a + (n-1)d, to find the first term aa and common difference dd.
4
Calculate the sum of the first 5 terms of the arithmetic progression
With a=4a = 4, d=5d = 5, and n=5n = 5, S5=52[2(4)+(5−1)5]=52[8+20]=52(28)=70S_5 = \frac{5}{2}[2(4) + (5-1)5] = \frac{5}{2}[8 + 20] = \frac{5}{2}(28) = 70.
Applying the AP sum formula Sn=n2[2a+(n−1)d]S_n = \frac{n}{2}[2a + (n-1)d].
5
Order the evaluated quantities from smallest to largest
Comparing numerical results 27<48<59<7027 < 48 < 59 < 70 gives the sequence: item 1, item 2, item 4, item 3.
Sorting the calculated values in ascending numerical order.

Key Concept

Evaluation and comparison of terms and sums in Arithmetic and Geometric Progressions
Estimated Time:2m 30s
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Topics

Arithmetic and Geometric Progressions (AP and GP)

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