Let a_1,a_2,a_3,. be in an arithmetic progression of positive terms. n Let A_k = a_1^2 - a_2^2 + a_3^2 - a_4^2 + .. + a_2k - 1^2 - a_2k^2. If A_3 = - 153,A_5 = - 435 and a_1^2 + a_2^2 + a_3^2 = 66, then a_17 - A_7 is equal to

a_1, a_2, a_3, is an A.P. Let d be the common difference aligned & a_2-a_1=a_3-a_2==a_2 k-a_2 k-1=d \\ & Now, a_1^2-a_2^2+a_3^2-a_4^2++a_2 k-1^2-a_2 k^2 \\ & =(a_1-a_2)(a_1+a

Let a_{1},a_{2},a_{3},. be in an arithmetic progression of positive terms. Let A_{k} = a_{1}^{2} - a_{2}^{2} + a_{3}^{2} - a_{4}^{2} + .. + a_{2k - 1}^{2} - a_{2k}^{2}. If A_{3} = - 153,A_{5} = - 435 and a_{1}^{2} + a_{2}^{2} + a_{3}^{2} = 66, then a_{17} - A_{7} is equal to : Sequence & Series (Mathematics)

$a_1, a_2, a_3, \ldots$ is an A.P. Let $d$ be the common difference $\begin{aligned} & \therefore a_2-a_1=a_3-a_2=\cdots=a_{2 k}-a_{2 k-1}=d \\ & \text { Now, } a_1^2-a_2^2+a_3^2-a_4^2+\cdots+a_{2 k-1}^2-a_{2 k}^2 \\ & =\left(a_1-a_2\right)\left(a_1+a_2\right)+\cdots+\left(a_{2 k-1}-a_{2 k}\right)\left(a_{2 k-1}+a_{2 k}\right) \\ & =-d\left[a_1+a_2+\cdots+a_{2 k-1}+a_{2 k}\right] \\ & =-d \cdot\left[\frac{2 k}{2}\left[a_1+a_{2 k}\right]\right]=-d k\left(a_1+a_{2 k}\right) \\ & \therefore A_k=-d k\left(a_1+a_{2 k}\right) \end{aligned}\\$ So, $A_3=-3 d\left(a_1+a_6\right)=-153$ $\\ \Rightarrow-3 d\left(a_1+a_1+5 d\right)=-153$ $\\ \Rightarrow-3 d\left(2 a_1+5 d\right)=-153 \\ \Rightarrow 2 a_1+5 d=\frac{51}{d}\\$ (i) Similarly $\\A_5=-435$ $\Rightarrow-5 d\left(2 a_1+9 d\right)$ =-435 $2a_1 + 9d = 87\\$ {d} (ii) Solving (i) and (ii) we get $4 d=\frac{36}{d} \Rightarrow d^2=9 \\ \Rightarrow d=3 \\$ [ ∵ Given A.P. is a progression of positive terms] $\\ \Rightarrow a_1=1 \\$ Now, $a_{17}=a_1+16 d=1+48 =49$ and $A_7=-3 \times 7\left(a_1+a_{14}\right)$ $\begin{aligned} & =-21(1+1+13 \times 3) \\&=-21(41) =-861 \\ & \therefore a_{17}-A_7=49+861=910\end{aligned}$

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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