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Zorluk: ZorArithmetic and Geometric Sequences and Series

Consider the four values defined below based on arithmetic and geometric sequences. Arrange these four items in ascending order (from smallest numerical value to largest numerical value).

  1. 1The 7th7\text{th} term of a geometric sequence with first term g1=3g_1 = 3 and common ratio r=2r = 2
  2. 2The sum of the first 88 terms of an arithmetic sequence with first term a1=5a_1 = 5 and common difference d=6d = 6
  3. 3The sum of an infinite geometric series with first term a1=150a_1 = 150 and common ratio r=13r = \frac{1}{3}
  4. 4The 50th50\text{th} term of an arithmetic sequence whose 3rd3\text{rd} term is 1515 and 7th7\text{th} term is 3535

Cevap

The correct ascending order from smallest to largest value is: (1) The 7th term of the geometric sequence (192192), (2) The sum of the first 8 terms of the arithmetic sequence (208208), (3) The sum of the infinite geometric series (225225), and (4) The 50th term of the arithmetic sequence (250250).
Evaluating each item yields: the 7th term of the geometric sequence equals 192, the sum of the first 8 terms of the arithmetic sequence equals 208, the sum of the infinite geometric series equals 225, and the 50th term of the arithmetic sequence equals 250. Placing these in ascending numerical order produces 192 < 208 < 225 < 250.

Adım Adım Çözüm

1
Calculate the value of the 7th term of the geometric sequence
Using g7=3⋅27−1=3⋅64=192g_7 = 3 \cdot 2^{7-1} = 3 \cdot 64 = 192.
The nthn\text{th} term of a geometric sequence is given by gn=g1rn−1g_n = g_1 r^{n-1}.
2
Calculate the sum of the first 8 terms of the arithmetic sequence
Using S8=82[2(5)+(8−1)6]=4[10+42]=208S_8 = \frac{8}{2}[2(5) + (8-1)6] = 4[10 + 42] = 208.
The sum of the first nn terms of an arithmetic sequence is given by Sn=n2[2a1+(n−1)d]S_n = \frac{n}{2}[2a_1 + (n-1)d].
3
Calculate the sum of the infinite geometric series
Using S∞=1501−1/3=1502/3=225S_\infty = \frac{150}{1 - 1/3} = \frac{150}{2/3} = 225.
The sum of an infinite geometric series with ∣r∣<1|r| < 1 is S∞=a11−rS_\infty = \frac{a_1}{1-r}.
4
Determine the common difference and the 50th term of the arithmetic sequence
Find d=35−157−3=5d = \frac{35 - 15}{7 - 3} = 5, then a50=15+(50−3)5=15+235=250a_{50} = 15 + (50 - 3)5 = 15 + 235 = 250.
Linear spacing between terms aka_k and ama_m yields am−ak=(m−k)da_m - a_k = (m - k)d.
5
Compare the calculated values to order them from smallest to largest
192<208<225<250192 < 208 < 225 < 250.
Arranging the quantities according to their numerical values gives the final sorted sequence.

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Arithmetic and Geometric Sequences and Series
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Arithmetic and Geometric Sequences and Series

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