The 7 th term of an AP is twice the third term. If the first term is 12 , find the common, difference.
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Correct Answer: Option A
Explanation:
To find the common difference of an arithmetic progression (AP) where the 7th term is twice the 3rd term, and the first term is 12, follow these steps:
1. Recall the formula for the \(n\)-th term of an AP:
\[
a_n = a + (n - 1) \cdot d
\]
where \(a\) is the first term and \(d\) is the common difference.
2. Write the expressions for the 7th term and the 3rd term:
- The 7th term (\(a_7\)) is:
\[
a_7 = a + 6d
\]
- The 3rd term (\(a_3\)) is:
\[
a_3 = a + 2d
\]
3. According to the problem, the 7th term is twice the 3rd term:
\[
a + 6d = 2(a + 2d)
\]
4. Substitute the first term \(a = 12\) into the equation:
\[
12 + 6d = 2(12 + 2d)
\]
5. Expand and simplify:
\[
12 + 6d = 24 + 4d
\]
\[
6d - 4d = 24 - 12
\]
\[
2d = 12
\]
\[
d = \frac{12}{2} = 6
\]
Thus, the common difference \(d\) is 6, which corresponds to option A.
To find the common difference of an arithmetic progression (AP) where the 7th term is twice the 3rd term, and the first term is 12, follow these steps:
1. Recall the formula for the \(n\)-th term of an AP:
\[
a_n = a + (n - 1) \cdot d
\]
where \(a\) is the first term and \(d\) is the common difference.
2. Write the expressions for the 7th term and the 3rd term:
- The 7th term (\(a_7\)) is:
\[
a_7 = a + 6d
\]
- The 3rd term (\(a_3\)) is:
\[
a_3 = a + 2d
\]
3. According to the problem, the 7th term is twice the 3rd term:
\[
a + 6d = 2(a + 2d)
\]
4. Substitute the first term \(a = 12\) into the equation:
\[
12 + 6d = 2(12 + 2d)
\]
5. Expand and simplify:
\[
12 + 6d = 24 + 4d
\]
\[
6d - 4d = 24 - 12
\]
\[
2d = 12
\]
\[
d = \frac{12}{2} = 6
\]
Thus, the common difference \(d\) is 6, which corresponds to option A.