Question:
If \(a_1, a_2, a_3, \ldots, a_n, \ldots\) are in A.P. such that \(a_4 - a_7 + a_{10} = m\), then the sum of first 13 terms of this A.P., is :
If \(a_1, a_2, a_3, \ldots, a_n, \ldots\) are in A.P. such that \(a_4 - a_7 + a_{10} = m\), then the sum of first 13 terms of this A.P., is :
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Write \(a_4 - a_7 + a_{10}\) in terms of \(a\) and \(d\); it simplifies to the middle term \(a_7\).
Updated On: Oct 31, 2026
- \(10\,m\)
- \(12\,m\)
- \(13\,m\)
- \(15\,m\)
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The Correct Option is C
Solution and Explanation
Let the first term be \(a\) and the common difference \(d\).
\(a_4 - a_7 + a_{10} = (a+3d) - (a+6d) + (a+9d) = a + 6d = a_7\).
So \(a_7 = m\).
\(S_{13} = \dfrac{13}{2}\,(2a + 12d) = 13(a + 6d) = 13\,a_7 = 13m\).
Answer: (3) \(13\,m\)
\(a_4 - a_7 + a_{10} = (a+3d) - (a+6d) + (a+9d) = a + 6d = a_7\).
So \(a_7 = m\).
\(S_{13} = \dfrac{13}{2}\,(2a + 12d) = 13(a + 6d) = 13\,a_7 = 13m\).
Answer: (3) \(13\,m\)
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