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For COMPETITION Number of Total Problems: 5. FOR PRINT ::: (Book)
The second and fourth terms of a geometric sequence are and . Which of the following is a possible first term?
Let the first term be and the common difference be . Therefore,
Dividing by eliminates the , yielding , so .
Now, since , , so .
We therefore see that is a possible first term.
Answer:
A sequence of three real numbers forms an arithmetic progression with a first term of . If is added to the second term and is added to the third term, the three resulting numbers form a geometric progression. What is the smallest possible value for the third term in the geometric progression?
Let be the common difference. Then are the terms of the geometric progression. Since the middle term is the geometric mean of the other two terms, . The smallest possible value occurs when , and the third term is .
Positive integers , , and , with , form a geometric sequence with an integer ratio. What is ?
The prime factorization of is . As , the ratio must be positive and larger than , hence there is only one possibility: the ratio must be , and then , and .
Consider the set of numbers . The ratio of the largest element of the set to the sum of the other ten elements of the set is closest to which integer?