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8. Sequences and Series
hard
माना $a_{1}, a_{2}, a_{3}, \ldots \ldots, a_{n}, \ldots .$ एक समांतर श्रेढ़ी में हैं। यदि $a_{3}+a_{7}+a_{11}+a_{15}=72$ है, तो उसके प्रथम $17$ पदों का योग बराबर है
A
$306$
B
$204$
C
$153$
D
$612$
(JEE MAIN-2016)
Solution
$\begin{array}{l}
{a_3} + {a_7} + {a_{11}} + {a_{15}} = 72\\
\left( {{a_3} + {a_{15}}} \right) + \left( {{a_7} + {a_{11}}} \right) = 72\\
{a_3} + {a_{15}} + {a_7} + {a_{11}} = \left( {{a_1} + {a_{17}}} \right)
\end{array}$
$\begin{array}{l}
{a_1} + {a_{17}} = 36\\
{S_{17}} = \frac{{17}}{2}\left[ {{a_1} + {a_{17}}} \right] = 17 \times 18 = 306
\end{array}$
Standard 11
Mathematics
