Final answer:
The number of values k can attain in the given series is 5. The correct answer is A. 5.
Step-by-step explanation:
The question asks to determine how many three-digit values of k are possible if the arithmetic mean (A.M.) and the geometric mean (G.M.) of the first and last numbers in a series, starting from 21 and ending at k, exist within the series. To find the A.M. of the first number, 21, and the last number, k, we use the formula:
A.M. = (21 + k)/2
Similarly, to find the G.M., we take the square root of their product:
G.M. = √(21 × k)
Since both A.M. and G.M. are part of the series (which comprises of consecutive natural numbers), they must also be whole numbers. For the A.M. to be a whole number, (21 + k) must be divisible by 2.
Considering that k is a three-digit number, its smallest possible value is 100 (as 99 would still be a two-digit number) and its highest possible value is 999. We must find k such that both A.M. and G.M. fall within the series, which constrains k to be between 100 and 999.
The series given is 21, 22, 23, ..., k - 1, k where k is a three-digit number. To find the number of values k can attain, we need to find the range of k.
The arithmetic mean (A.M.) of the series is the average of the first and last numbers, which is (21 + k)/2. The geometric mean (G.M.) is the square root of the product of all the numbers, which is √(21 * 22 * 23 * ... * k).
For the A.M. and G.M. to exist, we need (21 + k)/2 > 0 and √(21 * 22 * 23 * ... * k) > 0. From these inequalities, we can infer that k > -21 and k > 0. Therefore, the number of values k can attain is 5.