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Geometric Sequences: all problems

Focus mode: every practice problem from the chapter, one after another.

0 / 14 solved

Problem 16.1.1

MATHCOUNTS

The fifth term of a geometric sequence of positive numbers is 1111 and the eleventh term is 55. What is the eighth term of the sequence?

Problem 16.1.2

MATHCOUNTS

The third and fourth terms of an arithmetic sequence are the first and second terms of a geometric sequence. If the first two terms of the arithmetic sequence are 5,25, 2, then what is the fourth term of the geometric sequence?

Problem 16.1.3

MATHCOUNTS

The numbers aa, bb, cc, and dd form a geometric sequence, in that order. If bb is three more than aa, and cc is nine more than bb, what is the value of dd?

Problem 16.1.4

MATHCOUNTS

The positive integers AA, BB, and CC form an arithmetic sequence while the integers BB, CC, and DD form a geometric sequence. If CB=53\frac{C}{B} = \frac53, what is the smallest possible value of A+B+C+DA + B + C + D?

Problem 16.2.1(additional)

AIME

Call a 3-digit number geometric if it has 3 distinct digits which, when read from left to right, form a geometric sequence. Find the difference between the largest and smallest geometric numbers.

Problem 16.2.2(additional)

AIME

The sum of the first 20112011 terms of a geometric sequence is 200200. The sum of the first 40224022 terms is 380380. Find the sum of the first 60336033 terms.

Problem 16.2.3(additional)

AIME

Suppose nn is a positive integer and dd is a single digit in base 10. Find nn if

n810=0.d25d25d25…\frac{n}{810} = 0.d25d25d25\ldots

4 terms, g1 = 9/25, r = 1/100: g_4 = ≈ 3.600e-7, S_4 = ≈ 0.3636, S∞ = 4/11

first term g₁9/25
ratio r1/100
terms n4
Terms: 9/25, 9/2500, 9/250000, ≈ 3.600e-7
Running total after each term
limit
g4 = 9/25 × (1/100)3 = ≈ 3.600e-7
gn=g1⋅rn−1g_n = g_1 \cdot r^{n-1}
S4 = ≈ 0.3636
Sn=g1⋅1−rn1−rS_n = g_1 \cdot \frac{1 - r^n}{1 - r}
limit 4/11; gap left = ≈ 3.636e-9
S∞=g11−r=411S_\infty = \frac{g_1}{1-r} = \tfrac{4}{11}

r = 1/100: g_4 = ≈ 3.600e-7, S_4 = ≈ 0.3636, limit 4/11, gap ≈ 3.636e-9

Problem 16.2.4(additional)

AMC 10

The sum of an infinite geometric series is a positive number SS, and the second term in the series is 11. What is the smallest possible value of SS?

Problem 16.2.5(additional)

AMC 10

Positive integers aa, bb, and 20092009, with a<b<2009a < b < 2009, form a geometric sequence with an integer ratio. What is aa?

Problem 16.2.6(additional)

AIME

In an increasing sequence of four positive integers, the first three terms form an arithmetic progression, the last three terms form a geometric progression, and the first and fourth terms differ by 3030. Find the sum of the four terms.

Problem 16.2.7(additional)

AMC 12

A high school basketball game between the Raiders and Wildcats was tied at the end of the first quarter. The number of points scored by the Raiders in each of the four quarters formed an increasing geometric sequence, and the number of points scored by the Wildcats in each of the four quarters formed an increasing arithmetic sequence. At the end of the fourth quarter, the Raiders had won by one point. Neither team scored more than 100100 points. What was the total number of points scored by the two teams in the first half?

Problem 16.2.8(additional)

Let pp and qq be real numbers with ∣p∣<1|p| < 1 and ∣q∣<1|q| < 1 such that

p+pq+pq2+pq3+⋯=2andq+qp+qp2+qp3+⋯=3.p + pq + pq^2 + pq^3 + \cdots = 2 \quad\text{and}\quad q + qp + qp^2 + qp^3 + \cdots = 3.

Find 100pq100pq.

Problem 16.2.9(additional)

AIMEChallenge

A sequence of positive integers with a1=1a_1 = 1 and a9+a10=646a_9 + a_{10} = 646 is formed so that the first three terms are in geometric progression, the second, third, and fourth terms are in arithmetic progression, and, in general, for all n≥1n \ge 1, the terms a2n−1,a2n,a2n+1a_{2n-1}, a_{2n}, a_{2n+1} are in geometric progression, and the terms a2n,a2n+1a_{2n}, a_{2n+1}, and a2n+2a_{2n+2} are in arithmetic progression. Let ana_n be the greatest term in this sequence that is less than 10001000. Find n+ann + a_n.

Problem 16.2.10(additional)

AIMEChallenge

Two distinct, real, infinite geometric series each have a sum of 11 and have the same second term. The third term of one of the series is 1/81/8, and the second term of both series can be written in the form m−np\frac{\sqrt{m} - n}{p}, where mm, nn, and pp are positive integers and mm is not divisible by the square of any prime. Find 100m+10n+p100m + 10n + p.