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MathHardAdv 2020 • Paper 1

Number of Possible Values of First Term for Equality of Sequences

Let a1,a2,a3,…a_1, a_2, a_3, \dots be a sequence of positive integers forming an arithmetic progression with a common difference of 22. Further, let b1,b2,b3,…b_1, b_2, b_3, \dots be a sequence of positive integers forming a geometric progression with a common ratio of 22. If a1=b1=ca_1 = b_1 = c, then the total count of all possible values of cc, for which the equality 2(a1+a2+⋯+an)=b1+b2+⋯+bn2(a_1 + a_2 + \dots + a_n) = b_1 + b_2 + \dots + b_n holds for some positive integer nn, is _____

Official Numerical Answer1

Topics & Concepts

Sequences and SeriesArithmetic ProgressionGeometric ProgressionSum of Series

Step-by-Step Solution

The given sequence a1,a2,a3,…a_1, a_2, a_3, \dots forms an arithmetic progression (AP) with first term a1=ca_1 = c and common difference d=2d = 2. The sum of the first nn terms of this AP is given by: Sa,n=a1+a2+⋯+an=n2[2c+(n−1)2]=n(c+n−1)S_{a,n} = a_1 + a_2 + \dots + a_n = \frac{n}{2} \left[ 2c + (n - 1)2 \right] = n(c + n - 1)

The given sequence b1,b2,b3,…b_1, b_2, b_3, \dots forms a geometric progression (GP) with first term b1=cb_1 = c and common ratio r=2r = 2. The sum of the first nn terms of this GP is given by: Sb,n=b1+b2+⋯+bn=c(2n−1)S_{b,n} = b_1 + b_2 + \dots + b_n = c(2^n - 1)

We are given the relation: 2(a1+a2+⋯+an)=b1+b2+⋯+bn2(a_1 + a_2 + \dots + a_n) = b_1 + b_2 + \dots + b_n

Substituting the sum expressions into the equality: 2n(c+n−1)=c(2n−1)2n(c + n - 1) = c(2^n - 1)

Rearranging the terms to express cc in terms of nn: 2nc+2n(n−1)=c⋅2n−c2nc + 2n(n - 1) = c \cdot 2^n - c c(2n−2n−1)=2n(n−1)c(2^n - 2n - 1) = 2n(n - 1) c=2n(n−1)2n−2n−1c = \frac{2n(n - 1)}{2^n - 2n - 1}

Since cc must be a positive integer (c∈Z+c \in \mathbb{Z}^+), we analyze the values of cc for different positive integers nn:

  1. For n=1n = 1: c=2(1)(0)21−2(1)−1=0−1=0(Not a positive integer)c = \frac{2(1)(0)}{2^1 - 2(1) - 1} = \frac{0}{-1} = 0 \quad (\text{Not a positive integer})

  2. For n=2n = 2: c=2(2)(1)22−2(2)−1=4−1=−4(Not a positive integer)c = \frac{2(2)(1)}{2^2 - 2(2) - 1} = \frac{4}{-1} = -4 \quad (\text{Not a positive integer})

  3. For n=3n = 3: c=2(3)(2)23−2(3)−1=128−6−1=121=12(Valid positive integer)c = \frac{2(3)(2)}{2^3 - 2(3) - 1} = \frac{12}{8 - 6 - 1} = \frac{12}{1} = 12 \quad (\text{Valid positive integer})

  4. For n=4n = 4: c=2(4)(3)24−2(4)−1=2416−8−1=247(Not an integer)c = \frac{2(4)(3)}{2^4 - 2(4) - 1} = \frac{24}{16 - 8 - 1} = \frac{24}{7} \quad (\text{Not an integer})

  5. For n=5n = 5: c=2(5)(4)25−2(5)−1=4032−10−1=4021(Not an integer)c = \frac{2(5)(4)}{2^5 - 2(5) - 1} = \frac{40}{32 - 10 - 1} = \frac{40}{21} \quad (\text{Not an integer})

  6. For n=6n = 6: c=2(6)(5)26−2(6)−1=6064−12−1=6051=2017(Not an integer)c = \frac{2(6)(5)}{2^6 - 2(6) - 1} = \frac{60}{64 - 12 - 1} = \frac{60}{51} = \frac{20}{17} \quad (\text{Not an integer})

  7. For n≥7n \ge 7: Let us compare the numerator N(n)=2n(n−1)=2n2−2nN(n) = 2n(n-1) = 2n^2 - 2n and the denominator D(n)=2n−2n−1D(n) = 2^n - 2n - 1. For n=7n = 7: D(7)=27−2(7)−1=128−14−1=113D(7) = 2^7 - 2(7) - 1 = 128 - 14 - 1 = 113 N(7)=2(7)(6)=84N(7) = 2(7)(6) = 84 Since D(7)>N(7)>0D(7) > N(7) > 0, we have 0<c<10 < c < 1.

    By mathematical induction, for all n≥7n \ge 7, 2n>2n2+12^n > 2n^2 + 1, which implies D(n)>N(n)>0D(n) > N(n) > 0. Hence, 0<c<10 < c < 1 for all n≥7n \ge 7. Thus, cc can never be an integer for n≥7n \ge 7.

Therefore, the only possible value of cc is c=12c = 12 (corresponding to n=3n = 3).

The total count of all possible values of cc is 1.

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