Question

The value of x for which 2x, (x + 10) and (3x + 2) are the three consecutive terms of an A.P. is :

• A. 6
• B. - 6
• C. 18
• D. - 18

Hint:

For multiple-choice questions, you can also substitute the given options directly to find which one forms a valid A.P.
If you try option (A) \(x = 6\): the terms are \(12\), \(16\), and \(20\), which clearly have a common difference of 4.
This verification method is extremely useful for checking your answers!

Answer 1

The Correct Option is A

Step 1: Set up the terms using a common difference instead of the mean rule.
Let the common difference of the progression be $d$. Since the three given terms are consecutive terms of an AP, each term is the previous term plus $d$.
First term: $2x$
Second term: $2x + d = x + 10$
Third term: $2x + 2d = 3x + 2$

Step 2: Get $d$ from the first pair of terms.
From $2x + d = x + 10$:
\[ d = x + 10 - 2x = 10 - x \]

Step 3: Get $d$ again from the second pair of terms, and set the two expressions equal.
From $2x + 2d = 3x + 2$:
\[ 2d = 3x + 2 - 2x = x + 2 \]
\[ d = \frac{x + 2}{2} \]
Since both expressions equal the same common difference $d$:
\[ 10 - x = \frac{x + 2}{2} \]

Step 4: Solve the equation for $x$.
Multiply both sides by 2 to clear the fraction:
\[ 20 - 2x = x + 2 \]
\[ 20 - 2 = x + 2x \]
\[ 18 = 3x \]
\[ x = 6 \]
Checking, $d = 10 - 6 = 4$, and the terms become $12, 16, 20$, which indeed have a common difference of 4.

Final Answer:
The value of $x$ is 6, which corresponds to option (A). \[ \boxed{x = 6} \]