If a_1,a_2,a_3, ... ... ,a_n are in A.P. andn a_1 + a_4 + a_7 + ... ... + a_16 = 114, thenn a_1 + a_6 + a_11 + a_16 is equal to

nn n (b) : Given, a_1+a_4+a_7+a_10+a_13+a_16=114 n => (a_1+a_16)+(a_4+a_13)+(a_7+a_10)=114nn nnn nnn nn => 3(a_1+a_16)=114[Because a_1+a_16=a_4+a_13=a_7+a_10] nn => a_1+a_16=(114 / 3)=38 \ (i)nn nnn nNow, a_1+a_6+a_11+a_16=(a_1+a_16)+(a_6+a_11)nnnn n =2(a_1+a_16) n =2 x 38 n =76 n (Because a_1+a_16=a_6+a_11)nn nnnn[Using (i)]n

If a_{1},a_{2},a_{3},,a_{n} are in A.P. and a_{1} + a... | Sequence & Series (Mathematics)

\n\n\begin{aligned}\n& \text { (b) : Given, } a_1+a_4+a_7+a_{10}+a_{13}+a_{16}=114 \\\n& \Rightarrow\left(a_1+a_{16}\right)+\left(a_4+a_{13}\right)+\left(a_7+a_{10}\right)=114\n\n\end{aligned}\n\n

\n \n

\n\n\begin{gathered}\n\n\Rightarrow 3\left(a_1+a_{16}\right)=114\left[\because a_1+a_{16}=a_4+a_{13}=a_7+a_{10}\right] \\\n\n\Rightarrow a_1+a_{16}=\frac{114}{3}=38 \#(i)\n\n\end{gathered}\n\n

\n \nNow,

a_1+a_6+a_{11}+a_{16}=\left(a_1+a_{16}\right)+\left(a_6+a_{11}\right)

\n\n

\n\n\begin{aligned}\n& =2\left(a_1+a_{16}\right) \\\n& =2 \times 38 \\\n& =76 \\\n& \left(\because a_1+a_{16}=a_6+a_{11}\right)\n\n\end{aligned}\n\n

\n\n[Using (i)]\n

Explanation

Let a_{1},a_{2},,a_{n} be a given A.P. whose common difference is an integer and S_{n} = a_{1} + a_{2} + + a_{n}. If a_{1} = 1,a_{n} = 300 and 15 \leq n \leq 50, then the ordered pair ( S_{n - 4},a_{n - 4} ) is equal to

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Let \frac{1}{x_{1}},\frac{1}{x_{2}},,\frac{1}{x_{n}}( x_{i} eq 0 .\ for . \ i = 1,2,,n ) be in A.P. such that x_{1} = 4 and x_{21} = 20. If n is the least positive integer for which x_{n} > 50, then \sum_{i = 1}^{n}\mspace{2mu}( \frac{1}{x_{i}} ) is equal to

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