JEE Mains · Maths · STD 11 - 8. sequence and series
If \(a_1, a_2, a_3, …….\) are in \(A.P.\) such that \(a_1 + a_7 + a_{16} = 40\), then the sum of the first \(15\) terms of this \(A.P.\) is
- A \(200\)
- B \(280\)
- C \(150\)
- D \(120\)
Answer & Solution
Correct Answer
(A) \(200\)
Step-by-step Solution
Detailed explanation
\({a_1},{a_2},....{a_n}\) are in A.P. \({a_1} + {a_7} + {a_{16}} = 40\) \( \Rightarrow a + a + 6d + a + 15d = 40\) \( \Rightarrow 3a + 21d = 40\) \( \Rightarrow a + 7d = \frac{{40}}{3}\) \(515 = \frac{{15}}{2}\left[ {2a + 14d} \right]\) \( = 15\left[ {a + 7d} \right]\)…
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