Let $a_1, a_2, a_3, \ldots$ be in an arithmetic progression of positive terms.
Let $A_k=a_1^2-a_2^2+a_3^2-a_4^2+\ldots+a_{2 k-1}^2-a_{2 k}^2$.
If $\mathrm{A}_3=-153, \mathrm{~A}_5=-435$ and $\mathrm{a}_1^2+\mathrm{a}_2^2+\mathrm{a}_3^2=66$, then $\mathrm{a}_{17}-\mathrm{A}_7$ is equal to ________.
Answer (integer)
910
Solution
<p>Let $a_n=a+(n-1) d \forall n \in N$</p>
<p>$$\begin{aligned}
A_k & =\left(a_1^2-a_2^2\right)+\left(a_3^2-a_4^2\right)+\ldots a_{2 k-1}^2-a_{2 k}^2 \\
& =(-d)\left(a_1+a_2+\ldots+a_{2 k}\right) \\
& A_k=(-d k)(2 a+(2 k-1) d) \\
\Rightarrow & A_3=(-3 d)(2 a+5 d)=-153 \\
\Rightarrow & d(2 a+5 d)=51 \quad \text{... (i)}\\
& A_5=(-5 d)(2 a+9 d)=-435
\end{aligned}$$</p>
<p>$$\begin{aligned}
\Rightarrow & d(2 a+9 d)=87 \\
\Rightarrow & 4 d^2=36 \Rightarrow d= \pm 3(d=3 \text { positive terms }) \\
\Rightarrow & 3(2 a+27)=87 \\
\Rightarrow & 2 a=29-27 \\
\Rightarrow & a=1 \\
& a_{17}-A_7=(a+16 d)-(-7 d)(2+13 d) \\
& =49+7 \times 3(2+39) \\
& =49+21 \times 41=910
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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