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Zorluk: Çok zorArithmetic and Geometric Sequences and Series

An arithmetic sequence and a geometric sequence both have a first term equal to 22. The 3rd term of the arithmetic sequence is equal to the 2nd term of the geometric sequence, and the 7th term of the arithmetic sequence is equal to the 3rd term of the geometric sequence. If the common ratio of the geometric sequence is not equal to 11, what is the sum of the first 1010 terms of the arithmetic sequence?

  1. A
    20
  2. B
    30
  3. C
    54
  4. 65Cevap
  5. E
    130

Cevap

The sum of the first 10 terms of the arithmetic sequence is 65.
By writing the equations for the given sequence terms (2+2d=2r2 + 2d = 2r and 2+6d=2r22 + 6d = 2r^2), we substitute d=r−1d = r - 1 into the quadratic equation to find r2−3r+2=0r^2 - 3r + 2 = 0. Since r≠1r \neq 1, we find r=2r = 2, which gives a common difference d=1d = 1. Applying the sum formula for an arithmetic sequence for n=10n = 10 yields S10=102(2(2)+9(1))=5(13)=65S_{10} = \frac{10}{2}(2(2) + 9(1)) = 5(13) = 65.

Adım Adım Çözüm

1
Express the 3rd and 7th terms of the arithmetic sequence and 2nd and 3rd terms of the geometric sequence.
Arithmetic terms: a3=2+2da_3 = 2 + 2d and a7=2+6da_7 = 2 + 6d. Geometric terms: b2=2rb_2 = 2r and b3=2r2b_3 = 2r^2.
The nn-th term of an arithmetic sequence is an=a1+(n−1)da_n = a_1 + (n-1)d and of a geometric sequence is bn=b1rn−1b_n = b_1 r^{n-1}.
2
Set up the system of equations given by the problem.
2+2d=2r  ⟹  1+d=r  ⟹  d=r−12 + 2d = 2r \implies 1 + d = r \implies d = r - 1, and 2+6d=2r2  ⟹  1+3d=r22 + 6d = 2r^2 \implies 1 + 3d = r^2.
We are given a3=b2a_3 = b_2 and a7=b3a_7 = b_3 with a1=b1=2a_1 = b_1 = 2.
3
Substitute d=r−1d = r - 1 into the second equation and solve for rr.
1+3(r−1)=r2  ⟹  r2−3r+2=0  ⟹  (r−1)(r−2)=01 + 3(r - 1) = r^2 \implies r^2 - 3r + 2 = 0 \implies (r - 1)(r - 2) = 0. Since r≠1r \neq 1, r=2r = 2.
Solving the quadratic equation yields two roots, but the problem excludes r=1r = 1.
4
Determine the common difference dd and compute the sum of the first 10 terms of the arithmetic sequence.
d=2−1=1d = 2 - 1 = 1. S10=102[2(2)+(10−1)(1)]=5(4+9)=65S_{10} = \frac{10}{2}[2(2) + (10 - 1)(1)] = 5(4 + 9) = 65.
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].

Anahtar Kavram

Combining Arithmetic and Geometric Sequence Formulas
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Arithmetic and Geometric Sequences and Series

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