Math Problem Statement

Assertion: Let the positive numbers π‘Ž , 𝑏 , 𝑐 a,b,c be in Arithmetic Progression (A.P.). Then 1 𝑏 𝑐 , 1 π‘Ž 𝑐 , 1 π‘Ž 𝑏 bc 1 ​ , ac 1 ​ , ab 1 ​ are also in A.P.

Reason: If each term of an A.P. is divided by π‘Ž 𝑏 𝑐 abc, the resulting sequence is also in A.P.

Solution

Let's analyze the assertion and reason step-by-step.

Assertion:

We are given that aa, bb, and cc are positive numbers in Arithmetic Progression (A.P.). The claim is that the terms:

1bc,1ac,1ab\frac{1}{bc}, \frac{1}{ac}, \frac{1}{ab}

are also in A.P..

Reason:

We are asked to reason that if each term of an A.P. is divided by the product abcabc, the resulting sequence will also be in A.P..

Step-by-Step Solution:

1. Definition of Arithmetic Progression:

A sequence x1,x2,x3x_1, x_2, x_3 is in A.P. if the common difference between consecutive terms is constant, i.e.,

x2βˆ’x1=x3βˆ’x2x_2 - x_1 = x_3 - x_2

2. Given A.P.: a,b,ca, b, c are in A.P..

For aa, bb, and cc to be in A.P., the common difference condition must hold. Hence,

bβˆ’a=cβˆ’bor equivalently2b=a+cb - a = c - b \quad \text{or equivalently} \quad 2b = a + c

3. Sequence to Verify: 1bc,1ac,1ab\frac{1}{bc}, \frac{1}{ac}, \frac{1}{ab}.

Let us define these terms as x1=1bcx_1 = \frac{1}{bc}, x2=1acx_2 = \frac{1}{ac}, and x3=1abx_3 = \frac{1}{ab}.

We need to verify if the sequence x1,x2,x3x_1, x_2, x_3 satisfies the condition for an A.P.:

x2βˆ’x1=x3βˆ’x2x_2 - x_1 = x_3 - x_2

Let’s compute the differences:

x2βˆ’x1=1acβˆ’1bc=bβˆ’aabcx_2 - x_1 = \frac{1}{ac} - \frac{1}{bc} = \frac{b - a}{abc} x3βˆ’x2=1abβˆ’1ac=cβˆ’babcx_3 - x_2 = \frac{1}{ab} - \frac{1}{ac} = \frac{c - b}{abc}

Now, since bβˆ’a=cβˆ’bb - a = c - b (from the A.P. condition for a,b,ca, b, c), we have:

bβˆ’aabc=cβˆ’babc\frac{b - a}{abc} = \frac{c - b}{abc}

Thus,

x2βˆ’x1=x3βˆ’x2x_2 - x_1 = x_3 - x_2

Therefore, the sequence 1bc,1ac,1ab\frac{1}{bc}, \frac{1}{ac}, \frac{1}{ab} is indeed in A.P..

Conclusion:

The given assertion is correct. The sequence 1bc,1ac,1ab\frac{1}{bc}, \frac{1}{ac}, \frac{1}{ab} is in A.P. when a,b,ca, b, c are in A.P..


Would you like more details on any part of the solution or have additional questions on the topic? Here are some related questions you may find useful:

  1. What is the general form of an A.P. and how is the common difference calculated?
  2. Can you prove a sequence of fractions is in A.P. if their numerators are in A.P.?
  3. How does the harmonic progression relate to A.P.?
  4. What other types of progressions (like geometric progression) have similar properties?
  5. How can we generalize this result to more than three terms in A.P.?

Tip: In problems involving A.P., always check the condition 2b=a+c2b = a + c for three terms to confirm whether they follow an A.P.

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Math Problem Analysis

Mathematical Concepts

Arithmetic Progression (A.P.)
Sequences
Fractions

Formulas

x2 - x1 = x3 - x2 for A.P.
A.P. condition: 2b = a + c

Theorems

Properties of Arithmetic Progression

Suitable Grade Level

Grades 10-12