JEE Mains · Maths · STD 11 - 8. sequence and series
If \(a_1, a_2, a_3 …………\) an are in \(A.P\) and \(a_1 + a_4 + a_7 + …………… + a_{16} = 114\), then \(a_1 + a_6 + a_{11} + a_{16}\) is equal to
- A \(76\)
- B \(64\)
- C \(98\)
- D \(38\)
Answer & Solution
Correct Answer
(A) \(76\)
Step-by-step Solution
Detailed explanation
\({a_1} + {a_4} + {a_{ 7}} + {a_{10}} + {a_{13}} + {a_{16}} = 114\) \( \Rightarrow \frac{6}{2}\left( {{a_1} + {a_{16}}} \right) = 114\) \({a_1} + {a_{16}} = 38\) So, \({a_1} + {a_6} + {a_{ 11}} + {a_{16}} = \frac{4}{2}\left( {{a_1} + {a_{16}}} \right)\)…
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