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(b) Series, $2 + 5 + 8 + 11............$

$a=2,d=3$ and let number of terms is $n$

then sum of $A.P.$ ${a_1} + {a_5} + {a_{10}} + {a_{15}} + {a_{

Vedclass · Accepted Answer

$200$

Vedclass · Answer

(b) Series, $2 + 5 + 8 + 11............$

$a=2,d=3$ and let number of terms is $n$

then sum of $A.P.$ ${a_1} + {a_5} + {a_{10}} + {a_{15}} + {a_{20}} + {a_{24}} = 225$

$ \Rightarrow $ $60100 = \frac{n}{2}\left\{ {2 	imes 2 + (n - 1)3} ight\}$

$ \Rightarrow $ $120200 = n(3n + 1)$

$ \Rightarrow $ $3{n^2} + n - 120200 = 0$

$ \Rightarrow $ $(n - 200)(3n + 601) = 0$

Hence $n = 200$.

If the sum of the series $2 + 5 + 8 + 11............$ is $60100$, then the number of terms are

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