Question

If the numbers \( a_1, a_2, \ldots, a_n \) are different from zero and form an arithmetic progression, then \[ \frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} + ... + \frac{1}{a_n} \] is equal to

• A. \( \frac{1}{a_1 a_n} \)
• B. \( \frac{n}{a_1 a_n} \)
• C. \( \frac{n+1}{a_1 a_n} \)
• D. \( \frac{n-1}{a_1 a_n} \)

Hint:

When dealing with arithmetic progressions, utilize the properties of the progression to derive sums and relations between terms.

Answer 1

The Correct Option is D

Let the numbers \( a_1, a_2, \dots, a_n \) form an arithmetic progression, which means the difference between consecutive terms is constant.
The sum of the reciprocals of the terms in an arithmetic progression can be determined using the following steps: 1.
The sum of reciprocals for an arithmetic progression can be written as: \[ \frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_n} \] 2.
Using the properties of arithmetic progression, we know that this sum simplifies to: \[ \frac{a_1}{a_n} + 1 \]