It is known that \[\begin{align*} \dfrac{t_{11}+t_{13}}{ t_{5}+t_{7}}&=\dfrac{187500}{1500}\\ \displaystyle\dfrac{ar^{10}+ar^{12}}{ar^4+ar^6}&=125\\ \dfrac{ar^{10}(1+r^2)}{ar^4(1+r^2)}&=125\\ r^6&=125\end{align*}\] Thus, \(r=\pm\sqrt{5}\). Therefore, \(ar^4+ar^6=25a+125a =150a=1500\) and so \(a=10\). Therefore, the sequence begins \(10, 10\sqrt{5}, 50\) or \(10, -10\sqrt{5}, 50\).
Let \(d\) be the common difference in the arithmetic sequence. Since the sequence has distinct terms, we know that \(d \neq 0\). Then \(b-c=-d\), \(c-a=2d\) and \(a-b=-d\). Thus, \[\begin{align*} -dx^{2}+2dx-d&=0\\ -d(x-1)^{2}&=0\end{align*}\] and since \(d \ne 0\), we have \(x=1\).
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, \(xy = \dfrac{3}{r}(3r) = 9\). Thus, \(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{x+y}{xy}=\dfrac{8}{9}\).
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 \(\dfrac{a}{r}\), \(a\), and \(ar\). Thus, \(a^3=125\) and so \(a=5\). Therefore, the numbers are \(\dfrac{5}{r},5\), and \(5r\). Let \(d\) be the common difference of the arithmetic sequence. We know that \(\dfrac{5}{r}\) is the first term of the arithmetic sequence and 5 is the third term. Therefore, \(5-\dfrac{5}{r} = 2d\). We also know that \(5r\) is the sixth term of the arithmetic sequence and therefore, \(5r-5 = 3d\). Therefore, \[\begin{align*} \dfrac{5-\dfrac{5}{r}}{5r-5}&=\dfrac{2d}{3d}\\ 3\left(5-\dfrac{5}{r}\right)&=2(5r-5) \\ 3-\dfrac{3}{r}&=2r-2\\ 0&=2r^{2}-5r+3\\ 0&=(2r-3)(r-1)\end{align*}\] 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 \(r=\dfrac{3}{2}\) gives the sequence \(\dfrac{10}{3}, 5, \dfrac{15}{2}\).
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 \(\dfrac{a}{r}\), \(a\), 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 \(d=\dfrac{5}{6}\). 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 \(d=\dfrac{5}{6}\), the sequence is \(\dfrac{10}{3}, 5, \dfrac{15}{2}\).
Our sum is \[\begin{align*} \displaystyle\sum_{k=1}^{N}\dfrac{k^2+k}{2}&=\dfrac{\sum_{k=1}^{N}k^{2}+\sum_{k=1}^{N}k}{2}\\ &=\dfrac{\dfrac{N(N+1)(2N+1)}{6}+\dfrac{N(N+1)}{2}}{2}\\ &=\dfrac{N(N+1)}{4}\left(\dfrac{2N+1}{3}+1\right)\\ &=\dfrac{N(N+1)}{4} \left(\dfrac{2N+4}{3}\right)\\ &=\dfrac{N(N+1)(N+2)}{6}\end{align*}\] Taking \(N =200\) we get \[\begin{align*} \displaystyle\sum_{k=1}^{200}\dfrac{k^2+k}{2}&=\dfrac{200\cdot201\cdot202}{6}\\ &= 1\,353\,400.\end{align*}\]
Represent the angles as \(a-2d\), \(a-d\), \(a\), \(a+d\) and \(a+2d\). The sum of these values is \(540\degree\). Therefore, \(5a=540\degree\) and so \(a=108\degree\). So either \(a-d=90\degree\) or \(a-2d=90\degree\). So the largest angle is either \(126\degree\) or \(144\degree\).
We let the four positive integers be represented by \(k\), \(kr\), \(kr^2\) and \(kr^3\). Then \[\begin{align*} kr+kr^2&=30 \tag{1}\\ k+kr^3&=35 \tag{2}\end{align*}\] Dividing \((2)\) by \((1)\) gives \[\begin{align*} \dfrac{k+kr^3}{kr+kr^2}&=\dfrac{35}{30}\\ \dfrac{1+r^3}{r+r^2}&=\dfrac{7}{6} & \text{(\(k\neq 0\) since \(kr+ kr^2 =30\))}\\ 6r^3-7r^2-7r+6&=0\end{align*}\] 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 \(r=-1, \dfrac{2}{3}\) or \(\dfrac{3}{2}\)
Using \(r=-1\), equation \((2)\) gives \(0k=35\), which is impossible. (It would also violate the condition \(a < b < c < d\). )
Using \(r=\dfrac{2}{3}\) in \((1)\), we find \(k=27\).
Using \(r=\dfrac{3}{2}\) in \((1)\) we find \(k=8\).
Both of these value give the same list of numbers, and when arranged in increasing order they are \((a,b,c,d)=(8,12,18,27)\).
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 \(\dfrac{5}{27}\).
Solution 1
Since there are an odd number of integers, the average of the integers is the middle integer. Therefore, the middle integer is \(\dfrac{500}{25}=20\). 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 \(500 = \dfrac{25}{2}(a+ (a+24))\), which simplifies to \(40=2a+24\). Therefore, \(a = 8\),
The common difference is \(d=2\), the first term is \(a=-1994\) and so \[-1994+2(n-1) = -1994\] Solving for \(n\) gives \(n=1995\).
\(S_1=t_1=3^1-1=2\).
\(S_2=t_1+t_2=3^2-1 =8\) and so \(t_2=8-2=6\).
\(S_3=t_1+t_2+t_3=3^3-1=26\) and so
\(t_3=26-8=18\).
\[\begin{align*} \dfrac{t_{n+1}}{t_n}&=\dfrac{S_{n+1}-S_n}{S_n-S_{n-1}}\\ &=\dfrac{(3^{n+1}-1)-(3^n-1)}{(3^n-1)-(3^{n-1}-1)}\\ &=\dfrac{3^n\cdot(3-1)}{3^{n-1}\cdot(3-1)}\\ &=3.\end{align*}\]
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 \(28\,001\) is \(28\,000\) (We see that \(\dfrac{28\,001}{7} \approx 4000.1\) and \(7 \cdot 4000 =28\,000\).) So \(n-1=\dfrac{28\,000-(42)}{7}\) and \(n=3995\).
We know \(f(n+1)=f(n)+\dfrac{1}{3}\) 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 \(\dfrac{1}{3}\). Therefore, \(f(100)=2+99\left(\dfrac{1}{3}\right)=35\).
Substituting for \(x\) and \(y\), \(-p+2q=r\) so \(q-p=r-q\) and we are done!
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 \(d\neq 0\). For any three-term geometric sequence, \(x_1, x_2, x_3\) we have \(x_1x_3=(x_2)^2\). So \[\begin{align*} (a+4d)(a+15d)&=(a+8d)^2\\ a^2+19ad+60d^2&=a^2+16ad+64d^2\\ 3ad&=4d^2\\ d&=\dfrac{3}{4}a & \text{(Since $d \neq 0$)}\end{align*}\] Thus, the general term is \[\begin{align*} t_k&=a+(k-1)\dfrac{3}{4}a\\ &=\dfrac{a}{4}(3k+1)\end{align*}\]
Therefore, \[r=\dfrac{t_9}{t_5}=\dfrac{\dfrac a4(3\cdot 9+1)}{\dfrac a4(3\cdot 5+1)}=\dfrac 74\]
We need to find an infinite number of triples \((i, j, k)\) such that \[\dfrac{t_j}{t_i}=\dfrac{t_k}{t_j}=\dfrac 74\] which is to say that \[\dfrac{3j+1}{3i+1}=\dfrac{3k+1}{3j+1}=\dfrac 74\]
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 \[\begin{align*} 3i+1 & = 16n = 3(5n)+n \\ 3j+1 & = 28n = 3(9n)+n \\ 3k+1 & = 49 n = 3(16n) +n\end{align*}\]
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 \[\begin{align*} i&=\dfrac{16(3m+1)-1}{3}=16m+5\\ j&=\dfrac{28(3m+1)-1}{3}=28m+9\\ k&=\dfrac{49(3m+1)-1}3=49m+16\end{align*}\] For each value of \(m\) we will obtain a three-term geometric sequence with common ratio \(\dfrac{7}{4}\).
The sequence goes \(5, 3, -2, -5, -3, 2, 5, 3, \ldots\). The sequence repeats in groups of \(6\) whose sum is \(0\). So the sum of \(32\) terms is \(5+3=8\).
\[\begin{align*} t_{1998}&=\dfrac{1995}{1997}\times t_{1996} \\ &= \dfrac{1995}{1997} \times \dfrac{1997}{1995} \times t_{1994} \\ & = \dfrac{1995}{1997} \times \dfrac{1997}{1995} \times \dfrac{1995}{1993} \times \cdots \times \dfrac{3}{5} \times \dfrac{1}{3} \times t_2 \\ &= - \dfrac{1}{1997}\end{align*}\]
The first term is \(t_1 = 555-7= 548\) and the common difference is \(-7\). Therefore, the sum is \(S_n=\dfrac{n}{2}[2(548)+(n-1)(-7)]\). 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 \(n=158\).