Sequences and Series
Solutions

1. It is known that $$\begin{array}{rl}{\displaystyle \frac{t_{11}+t_{13}}{t_5+t_7}}& ={\displaystyle \frac{187500}{1500}}\\ {\displaystyle {\displaystyle \frac{a{r}^{10}+a{r}^{12}}{a{r}^4+a{r}^6}}}& =125\\ {\displaystyle \frac{a{r}^{10}(1+r^2)}{a{r}^4(1+r^2)}}& =125\\ r^6& =125\end{array}$$ Thus, $r=\pm \sqrt{5}$. Therefore, $a{r}^4+a{r}^6=25a+125a=150a=1500$ and so $a=10$. Therefore, the sequence begins $10,10\sqrt{5},50$ or $10,-10\sqrt{5},50$.

2. Let $d$ be the common difference in the arithmetic sequence. Since the sequence has distinct terms, we know that $d\ne 0$. Then $b-c=-d$, $c-a=2d$ and $a-b=-d$. Thus, $$\begin{array}{rl}-d{x}^2+2dx-d& =0\\ -d(x-1{)}^2& =0\end{array}$$ and since $d\ne 0$, we have $x=1$.

3. From the arithmetic sequence we have that $4=x+d$ and $y=4+d$, where $d$ is the common difference. Therefore, $x+y=4-d+4+d=8$. From the geometric sequence we have $xr=3$ and $3r=y$, where $r$ is the common ratio (which is not $0$ since the second term is $3$). Therefore, $xy={\displaystyle \frac{3}{r}}(3r)=9$. Thus, ${\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{y}}={\displaystyle \frac{x+y}{xy}}={\displaystyle \frac{8}{9}}$.

4. Solution 1

   Since the product of the three numbers is non-zero, so is $r$, the common ratio of the geometric sequence. We let the numbers be ${\displaystyle \frac{a}{r}}$, $a$, and $ar$. Thus, $a^3=125$ and so $a=5$. Therefore, the numbers are ${\displaystyle \frac{5}{r}},5$, and $5r$. Let $d$ be the common difference of the arithmetic sequence. We know that ${\displaystyle \frac{5}{r}}$ is the first term of the arithmetic sequence and 5 is the third term. Therefore, $5-{\displaystyle \frac{5}{r}}=2d$. We also know that $5r$ is the sixth term of the arithmetic sequence and therefore, $5r-5=3d$. Therefore, $$\begin{array}{rl}{\displaystyle \frac{5-{\displaystyle \frac{5}{r}}}{5r-5}}& ={\displaystyle \frac{2d}{3d}}\\ 3(5-{\displaystyle \frac{5}{r}})& =2(5r-5)\\ 3-{\displaystyle \frac{3}{r}}& =2r-2\\ 0& =2r^2-5r+3\\ 0& =(2r-3)(r-1)\end{array}$$ The solution $r=1$ gives the sequence $5,5,5$, but we were told that the three numbers are distinct and so we discard this solution. The solution $r={\displaystyle \frac{3}{2}}$ gives the sequence ${\displaystyle \frac{10}{3}},5,{\displaystyle \frac{15}{2}}$.

   Solution 2

   Since the product of the three numbers is non-zero, so is $r$, the common ratio of the geometric sequence. We let the numbers be ${\displaystyle \frac{a}{r}}$, $a$, and $ar$. Thus, $a^3=125$ and so $a=5$. We know that the middle term, $5$, is the third term of an arithmetic sequence.

   Let $d$ be the common difference of this arithmetic sequence. The first term is the first term of the arithmetic sequence and therefore, it is $5-2d$. The third term is the sixth term of the arithmetic sequence and therefore, it is $5+3d$. The product of these three terms is $125$ and so $(5-2d)5(5+3d)=125$. Dividing both sides by $5$ and simplifying gives $-6d^2+5d=0$. Therefore, $d=0$ or $d={\displaystyle \frac{5}{6}}$. If $d=0$, then the sequence is $5,5,5$, but we were told that the three numbers are distinct and so we discard this solution. If $d={\displaystyle \frac{5}{6}}$, the sequence is ${\displaystyle \frac{10}{3}},5,{\displaystyle \frac{15}{2}}$.

5. Our sum is $$\begin{array}{rl}{\displaystyle \sum _{k=1}^N{\displaystyle \frac{k^2+k}{2}}}& ={\displaystyle \frac{\sum _{k=1}^N{k}^2+\sum _{k=1}^{N}k}{2}}\\ & ={\displaystyle \frac{{\displaystyle \frac{N(N+1)(2N+1)}{6}}+{\displaystyle \frac{N(N+1)}{2}}}{2}}\\ & ={\displaystyle \frac{N(N+1)}{4}}({\displaystyle \frac{2N+1}{3}}+1)\\ & ={\displaystyle \frac{N(N+1)}{4}}\left({\displaystyle \frac{2N+4}{3}}\right)\\ & ={\displaystyle \frac{N(N+1)(N+2)}{6}}\end{array}$$ Taking $N=200$ we get $$\begin{array}{rl}{\displaystyle \sum _{k=1}^{200}{\displaystyle \frac{k^2+k}{2}}}& ={\displaystyle \frac{200\cdot 201\cdot 202}{6}}\\ & =1353400.\end{array}$$

6. Represent the angles as $a-2d$, $a-d$, $a$, $a+d$ and $a+2d$. The sum of these values is $540\mathrm{°}$. Therefore, $5a=540\mathrm{°}$ and so $a=108\mathrm{°}$. So either $a-d=90\mathrm{°}$ or $a-2d=90\mathrm{°}$. So the largest angle is either $126\mathrm{°}$ or $144\mathrm{°}$.

7. We let the four positive integers be represented by $k$, $kr$, $k{r}^2$ and $k{r}^3$. Then $$\begin{array}{}\text{(1)}& kr+k{r}^2& =30\text{(2)}& k+k{r}^3& =35\end{array}$$ Dividing $(2)$ by $(1)$ gives $$\begin{array}{rl}{\displaystyle \frac{k+k{r}^3}{kr+k{r}^2}}& ={\displaystyle \frac{35}{30}}\\ {\displaystyle \frac{1+r^3}{r+r^2}}& ={\displaystyle \frac{7}{6}}& \text{(}k\ne 0\text{ since }kr+k{r}^2=30\text{)}\\ 6r^3-7r^2-7r+6& =0\end{array}$$ By inspection, we find that $r=-1$ is a solution. Using the factor theorem and long division, we arrive at $$(r+1)(2r-3)(3r-2)=0$$ So $r=-1,{\displaystyle \frac{2}{3}}$ or ${\displaystyle \frac{3}{2}}$

   Using $r=-1$, equation $(2)$ gives $0k=35$, which is impossible. (It would also violate the condition $a<b<c<d$. )

   Using $r={\displaystyle \frac{2}{3}}$ in $(1)$, we find $k=27$.

   Using $r={\displaystyle \frac{3}{2}}$ in $(1)$ we find $k=8$.

   Both of these value give the same list of numbers, and when arranged in increasing order they are $(a,b,c,d)=(8,12,18,27)$.

8. The sequence is arithmetic if and only if $t_1+t_3=2t_2$. There are $27$ equally likely ways to pick three numbers, of which only five lead to such a sequence:

   • $1,4,7$

   • $1,5,9$

   • $2,5,8$

   • $3,5,7$

   • $3,6,9$

   So the probability is ${\displaystyle \frac{5}{27}}$.

9. Solution 1

   Since there are an odd number of integers, the average of the integers is the middle integer. Therefore, the middle integer is ${\displaystyle \frac{500}{25}}=20$. Thus, the smallest integer is $8$.

   Solution 2

   The common difference is $1$ and the number of terms is $25$. Therefore, using the sum of an arithmetic sequence we get $500={\displaystyle \frac{25}{2}}(a+(a+24))$, which simplifies to $40=2a+24$. Therefore, $a=8$,

10. The common difference is $d=2$, the first term is $a=-1994$ and so $$-1994+2(n-1)=-1994$$ Solving for $n$ gives $n=1995$.

11. 1. $S_1=t_1=3^1-1=2$.
       $S_2=t_1+t_2=3^2-1=8$ and so $t_2=8-2=6$.
       $S_3=t_1+t_2+t_3=3^3-1=26$ and so $t_3=26-8=18$.

    2. $$\begin{array}{rl}{\displaystyle \frac{t_{n+1}}{t_n}}& ={\displaystyle \frac{S_{n+1}-S_n}{S_n-S_{n-1}}}\\ & ={\displaystyle \frac{(3^{n+1}-1)-(3^n-1)}{(3^n-1)-(3^{n-1}-1)}}\\ & ={\displaystyle \frac{3^n\cdot (3-1)}{3^{n-1}\cdot (3-1)}}\\ & =3.\end{array}$$

12. We can see that the $n$th term of the sequence is $7n$. The smallest multiple of $7$ that is greater than $40$ is $42$ and the largest multiple of $7$ that is less than $28001$ is $28000$ (We see that ${\displaystyle \frac{28001}{7}}\approx 4000.1$ and $7\cdot 4000=28000$.) So $n-1={\displaystyle \frac{28000-(42)}{7}}$ and $n=3995$.

13. We know $f(n+1)=f(n)+{\displaystyle \frac{1}{3}}$ and so the the function evaluated at positive integers gives a sequence that is arithmetic. Its first term is $2$ and its common difference is ${\displaystyle \frac{1}{3}}$. Therefore, $f(100)=2+99\left({\displaystyle \frac{1}{3}}\right)=35$.

14. Substituting for $x$ and $y$, $-p+2q=r$ so $q-p=r-q$ and we are done!

15. If the common difference is $0$, then the sequence is also a geometric sequence with a common ratio of $1$. In this case, any three terms form a three-term geometric sequence.

    Let’s consider what happens when $d\ne 0$. For any three-term geometric sequence, $x_1,x_2,x_3$ we have $x_1{x}_3=(x_2{)}^2$. So $$\begin{array}{rl}(a+4d)(a+15d)& =(a+8d{)}^2\\ a^2+19ad+60d^2& =a^2+16ad+64d^2\\ 3ad& =4d^2\\ d& ={\displaystyle \frac{3}{4}}a& \text{(Since }d\ne 0\text{)}\end{array}$$ Thus, the general term is $$\begin{array}{rl}{t}_k& =a+(k-1){\displaystyle \frac{3}{4}}a\\ & ={\displaystyle \frac{a}{4}}(3k+1)\end{array}$$

    Therefore, $$r={\displaystyle \frac{t_9}{t_5}}={\displaystyle \frac{{\displaystyle \frac{a}{4}}(3\cdot 9+1)}{{\displaystyle \frac{a}{4}}(3\cdot 5+1)}}={\displaystyle \frac{7}{4}}$$

    We need to find an infinite number of triples $(i,j,k)$ such that $${\displaystyle \frac{t_j}{t_i}}={\displaystyle \frac{t_k}{t_j}}={\displaystyle \frac{7}{4}}$$ which is to say that $${\displaystyle \frac{3j+1}{3i+1}}={\displaystyle \frac{3k+1}{3j+1}}={\displaystyle \frac{7}{4}}$$

    Therefore, $4(3j+1)=7(3i+1)$, which implies that $3j+1$ is a multiple of $7$ and $3i+1$ is a multiple of $4$.

    Also, $4(3k+1)=7(3j+1)$, which implies that $3k+1$ is a multiple of $7$ and $3j+1$ is a multiple of $4$. So $3j+1$ must be a multiple of $28$. Let $3j+1=28n$ for some integer $n$.

    We also have that $(3j+1{)}^2=(3i+1)(3k+1)$ and so $(3i+1)(3k+1)$ must be a multiple of ${28}^2$. So if we make $3i+1=16n$ and $3k+1=49n$, then we will have satisfied all the conditions.

    However, we need to guarantee that $i,j$ and $k$ are positive integers. We note that $$\begin{array}{rl}3i+1& =16n=3(5n)+n\\ 3j+1& =28n=3(9n)+n\\ 3k+1& =49n=3(16n)+n\end{array}$$

    So if we choose $n$ such that it is $1$ more than a multiple of $3$, then $i,j$ and $k$ will be integers. Therefore, let $n=3m+1$ for some non-negative integer $m$ and we obtain $$\begin{array}{rl}i& ={\displaystyle \frac{16(3m+1)-1}{3}}=16m+5\\ j& ={\displaystyle \frac{28(3m+1)-1}{3}}=28m+9\\ k& ={\displaystyle \frac{49(3m+1)-1}{3}}=49m+16\end{array}$$ For each value of $m$ we will obtain a three-term geometric sequence with common ratio ${\displaystyle \frac{7}{4}}$.

16. The sequence goes $5,3,-2,-5,-3,2,5,3,\dots$. The sequence repeats in groups of $6$ whose sum is $0$. So the sum of $32$ terms is $5+3=8$.

17. 
    

    $$\begin{array}{rl}{t}_{1998}& ={\displaystyle \frac{1995}{1997}}\times t_{1996}\\ & ={\displaystyle \frac{1995}{1997}}\times {\displaystyle \frac{1997}{1995}}\times t_{1994}\\ & ={\displaystyle \frac{1995}{1997}}\times {\displaystyle \frac{1997}{1995}}\times {\displaystyle \frac{1995}{1993}}\times \cdots \times {\displaystyle \frac{3}{5}}\times {\displaystyle \frac{1}{3}}\times t_2\\ & =-{\displaystyle \frac{1}{1997}}\end{array}$$

18. The first term is $t_1=555-7=548$ and the common difference is $-7$. Therefore, the sum is $S_n={\displaystyle \frac{n}{2}}[2(548)+(n-1)(-7)]$. Thus, the sum is negative when $1096+(n-1)(-7)<0$. Solving the equality $1096-7n+7=0$ we obtain $n\approx 157.6$. We note that $S_{157}=314$ and $S_{158}=-237$ Therefore, the smallest value of $n$ for which $S_n$ is negative is $n=158$.