The terms ${{u}_{1}}$, ${{u}_{2}}$, ${{u}_{3}}$, $\ldots $ form a geometric sequence with fi

Question

The terms ${{u}_{1}}$, ${{u}_{2}}$, ${{u}_{3}}$, $\ldots $ form a geometric sequence with first term $a$ and common ratio $r$, where $a\ne 0$. It is given that the sum of the first $10$ terms is $33$ times the sum of the first $5$ terms. Explain why $r$ cannot be equal to $1$.[1] Show that $r=2$.[2] If ${{u}_{7}}$ is the first term in the sequence greater than $11$, find exactly the range of possible values of $a$.[2] Given that $\sum\limits_{r=1}^{n}{\frac{1}{\left( 2r-1 \right)\left( 2r+1 \right)}=\frac{1}{2}-\frac{1}{4n+2}}$, find $\sum\limits_{r=4}^{n}{\frac{1}{\left( 2r+3 \right)\left( 2r+5 \right)}}$ in terms of $n$.[3]

Accepted Answer

(a)(i) If $r=1$, all the terms of the sequence will be equal to the first term $a$. Then the sum of the first $10$ terms will be $10a$ , which is only $2$ times the sum of the first $5$ terms $5a$ . (iii) $\frac{11}{64}\le a\le \frac{11}{32}$ (b) $\frac{1}{22}-\frac{1}{4n+10}$

2024 EJC Promo Q7

2024 EJC Promo Q7

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