118888
In a bank, the principal increases continuously at the rate of \(6 \%\) per year. Then the time required to doubling \(₹ \mathbf{6 0 0 0}\) rupees is........ (in year)
1 \(\frac{50}{3} \log 2\)
2 \(\frac{30}{3} \log 6\)
3 \(\frac{50}{3} \log 3\)
4 \(\frac{50}{3} \log 12\)
Explanation:
A According to question, let \(\mathrm{P}\) and \(\mathrm{t}\) be the principle and time. \(P=6000\) rupees getting double in time ' \(t\) ' with rate of \(6 \%\) per year, So \(\frac{\mathrm{dP}}{\mathrm{dt}}=\frac{6}{100} \mathrm{P}\) \(\int_{6000}^{12000} \frac{\mathrm{dP}}{\mathrm{P}}=\frac{3}{50} \int_0^{\mathrm{t}} \mathrm{dt}\) \({[\log \mathrm{P}]_{6000}^{12000}=\frac{3}{50} \mathrm{t}}\) \(\log 12000-\log 6000=\frac{3}{50} \mathrm{t}\) \(\log 2=\frac{3}{50} \mathrm{t}\)Time, \(t=\frac{50}{3} \log 2\)
AP EAMCET-21.09.2020
Sequence and Series
118889
Sum of the series \(1(1)+2(1+3)+3(1+3+5)\) \(+4(1+3+5+7)+\ldots .+10(1+3+5+7\) \(+\ldots .+19)\) is equal to
1 385
2 1025
3 1125
4 2025
5 3025
Explanation:
E Here, we have \(1(1)+2(1+3)+3(1+3+5)+4(1+3+5+7)+\ldots .\) \(+10(1+3+5+7+\ldots .+19)\) \(=1 \times 1+2 \times 2^2+3 \times 3^2+4 \times 4^2+\ldots . .+10 \times 10^2\) \(=\left[\frac{10(10+1)}{2}\right]^2 \quad\left[\begin{array}{l}\because \text { Sum of cubes of } \mathrm{n} \text { natural } \\ \text { numbers }=\left[\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right]^2\end{array}\right]\) \(=\left[\frac{10 \times 11}{2}\right]^2=(55)^2=3025\)
Kerala CEE-2016
Sequence and Series
118890
The value of \(\frac{1}{\sqrt{10}-\sqrt{9}}-\frac{1}{\sqrt{11}-\sqrt{10}}+\frac{1}{\sqrt{12}-\sqrt{11}}-\ldots .-\frac{1}{\sqrt{121}-\sqrt{120}}\) is equal to
118891
The sum of the series \(\frac{1^2}{1.2}+\frac{1^2+2^2}{2.3}+\frac{1^2+2^2+3^2}{3.4}+\ldots\). upto 20 terms is
1 \(\frac{205}{3}\)
2 \(\frac{200}{3}\)
3 \(\frac{220}{3}\)
4 \(\frac{210}{3}\)
Explanation:
C Let, \(S=\frac{1^2}{1.2}+\frac{1^2+2^2}{2.3}+\frac{1^2+2^2+3^2}{3.4}+\ldots\) upto 20 terms Let \(t_n\) be \(n^{\text {th }}\) term of series. Then, \(\mathrm{t}_{\mathrm{n}}=\frac{1^2+2^2+3^2+\ldots . .+\mathrm{n}^2}{\mathrm{n}(\mathrm{n}+1)}=\frac{2 \mathrm{n}+1}{6}\) Taking summation on both sides, we get- \(\sum_{\mathrm{n}=1}^{20} \mathrm{t}_{\mathrm{n}} =\frac{2}{6} \sum_{\mathrm{n}=1}^{20} \mathrm{n}+\frac{1}{6} \sum_{\mathrm{n}=1}^{20} 1\) \(=\frac{1}{3} \times\left(\frac{20(20+1)}{2}\right)+\frac{1}{6} \times 20\) \(\sum_{\mathrm{n}=1}^{20} \mathrm{t}_{\mathrm{n}} =\frac{1}{3} \times\left(\frac{20 \times 21}{2}\right)+\left(\frac{10}{3}\right)=\frac{210}{3}+\frac{10}{3}=\frac{220}{3}\)
COMEDK-2014
Sequence and Series
118892
\(11^3+12^3+13^3+\ldots \ldots \ldots \ldots . .+20^3\) is
1 an even integer
2 an odd integer divisible by 5
3 multiple of 10
4 an odd integer but not a multiple of 5
Explanation:
B Given that,\(11^3+12^3+13^3+\ldots .+20^3\) \(=\left(1^3+2^3+\ldots .+10^3+11^3+12^3+\ldots .+20^3\right)-\left(1^3+2^3+\ldots .+10^3\right)\) \(=\sum(20)^3-\sum(10)^3\) \(=\left(\frac{20(20+1)}{2}\right)^2-\left(\frac{10(10+1)}{2}\right)^2\) \(=(10 \times 21)^2-(5 \times 11)^2=(210-55)(210+55)\) \(=155 \times 265=41075=5 \times 8215\)
118888
In a bank, the principal increases continuously at the rate of \(6 \%\) per year. Then the time required to doubling \(₹ \mathbf{6 0 0 0}\) rupees is........ (in year)
1 \(\frac{50}{3} \log 2\)
2 \(\frac{30}{3} \log 6\)
3 \(\frac{50}{3} \log 3\)
4 \(\frac{50}{3} \log 12\)
Explanation:
A According to question, let \(\mathrm{P}\) and \(\mathrm{t}\) be the principle and time. \(P=6000\) rupees getting double in time ' \(t\) ' with rate of \(6 \%\) per year, So \(\frac{\mathrm{dP}}{\mathrm{dt}}=\frac{6}{100} \mathrm{P}\) \(\int_{6000}^{12000} \frac{\mathrm{dP}}{\mathrm{P}}=\frac{3}{50} \int_0^{\mathrm{t}} \mathrm{dt}\) \({[\log \mathrm{P}]_{6000}^{12000}=\frac{3}{50} \mathrm{t}}\) \(\log 12000-\log 6000=\frac{3}{50} \mathrm{t}\) \(\log 2=\frac{3}{50} \mathrm{t}\)Time, \(t=\frac{50}{3} \log 2\)
AP EAMCET-21.09.2020
Sequence and Series
118889
Sum of the series \(1(1)+2(1+3)+3(1+3+5)\) \(+4(1+3+5+7)+\ldots .+10(1+3+5+7\) \(+\ldots .+19)\) is equal to
1 385
2 1025
3 1125
4 2025
5 3025
Explanation:
E Here, we have \(1(1)+2(1+3)+3(1+3+5)+4(1+3+5+7)+\ldots .\) \(+10(1+3+5+7+\ldots .+19)\) \(=1 \times 1+2 \times 2^2+3 \times 3^2+4 \times 4^2+\ldots . .+10 \times 10^2\) \(=\left[\frac{10(10+1)}{2}\right]^2 \quad\left[\begin{array}{l}\because \text { Sum of cubes of } \mathrm{n} \text { natural } \\ \text { numbers }=\left[\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right]^2\end{array}\right]\) \(=\left[\frac{10 \times 11}{2}\right]^2=(55)^2=3025\)
Kerala CEE-2016
Sequence and Series
118890
The value of \(\frac{1}{\sqrt{10}-\sqrt{9}}-\frac{1}{\sqrt{11}-\sqrt{10}}+\frac{1}{\sqrt{12}-\sqrt{11}}-\ldots .-\frac{1}{\sqrt{121}-\sqrt{120}}\) is equal to
118891
The sum of the series \(\frac{1^2}{1.2}+\frac{1^2+2^2}{2.3}+\frac{1^2+2^2+3^2}{3.4}+\ldots\). upto 20 terms is
1 \(\frac{205}{3}\)
2 \(\frac{200}{3}\)
3 \(\frac{220}{3}\)
4 \(\frac{210}{3}\)
Explanation:
C Let, \(S=\frac{1^2}{1.2}+\frac{1^2+2^2}{2.3}+\frac{1^2+2^2+3^2}{3.4}+\ldots\) upto 20 terms Let \(t_n\) be \(n^{\text {th }}\) term of series. Then, \(\mathrm{t}_{\mathrm{n}}=\frac{1^2+2^2+3^2+\ldots . .+\mathrm{n}^2}{\mathrm{n}(\mathrm{n}+1)}=\frac{2 \mathrm{n}+1}{6}\) Taking summation on both sides, we get- \(\sum_{\mathrm{n}=1}^{20} \mathrm{t}_{\mathrm{n}} =\frac{2}{6} \sum_{\mathrm{n}=1}^{20} \mathrm{n}+\frac{1}{6} \sum_{\mathrm{n}=1}^{20} 1\) \(=\frac{1}{3} \times\left(\frac{20(20+1)}{2}\right)+\frac{1}{6} \times 20\) \(\sum_{\mathrm{n}=1}^{20} \mathrm{t}_{\mathrm{n}} =\frac{1}{3} \times\left(\frac{20 \times 21}{2}\right)+\left(\frac{10}{3}\right)=\frac{210}{3}+\frac{10}{3}=\frac{220}{3}\)
COMEDK-2014
Sequence and Series
118892
\(11^3+12^3+13^3+\ldots \ldots \ldots \ldots . .+20^3\) is
1 an even integer
2 an odd integer divisible by 5
3 multiple of 10
4 an odd integer but not a multiple of 5
Explanation:
B Given that,\(11^3+12^3+13^3+\ldots .+20^3\) \(=\left(1^3+2^3+\ldots .+10^3+11^3+12^3+\ldots .+20^3\right)-\left(1^3+2^3+\ldots .+10^3\right)\) \(=\sum(20)^3-\sum(10)^3\) \(=\left(\frac{20(20+1)}{2}\right)^2-\left(\frac{10(10+1)}{2}\right)^2\) \(=(10 \times 21)^2-(5 \times 11)^2=(210-55)(210+55)\) \(=155 \times 265=41075=5 \times 8215\)
118888
In a bank, the principal increases continuously at the rate of \(6 \%\) per year. Then the time required to doubling \(₹ \mathbf{6 0 0 0}\) rupees is........ (in year)
1 \(\frac{50}{3} \log 2\)
2 \(\frac{30}{3} \log 6\)
3 \(\frac{50}{3} \log 3\)
4 \(\frac{50}{3} \log 12\)
Explanation:
A According to question, let \(\mathrm{P}\) and \(\mathrm{t}\) be the principle and time. \(P=6000\) rupees getting double in time ' \(t\) ' with rate of \(6 \%\) per year, So \(\frac{\mathrm{dP}}{\mathrm{dt}}=\frac{6}{100} \mathrm{P}\) \(\int_{6000}^{12000} \frac{\mathrm{dP}}{\mathrm{P}}=\frac{3}{50} \int_0^{\mathrm{t}} \mathrm{dt}\) \({[\log \mathrm{P}]_{6000}^{12000}=\frac{3}{50} \mathrm{t}}\) \(\log 12000-\log 6000=\frac{3}{50} \mathrm{t}\) \(\log 2=\frac{3}{50} \mathrm{t}\)Time, \(t=\frac{50}{3} \log 2\)
AP EAMCET-21.09.2020
Sequence and Series
118889
Sum of the series \(1(1)+2(1+3)+3(1+3+5)\) \(+4(1+3+5+7)+\ldots .+10(1+3+5+7\) \(+\ldots .+19)\) is equal to
1 385
2 1025
3 1125
4 2025
5 3025
Explanation:
E Here, we have \(1(1)+2(1+3)+3(1+3+5)+4(1+3+5+7)+\ldots .\) \(+10(1+3+5+7+\ldots .+19)\) \(=1 \times 1+2 \times 2^2+3 \times 3^2+4 \times 4^2+\ldots . .+10 \times 10^2\) \(=\left[\frac{10(10+1)}{2}\right]^2 \quad\left[\begin{array}{l}\because \text { Sum of cubes of } \mathrm{n} \text { natural } \\ \text { numbers }=\left[\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right]^2\end{array}\right]\) \(=\left[\frac{10 \times 11}{2}\right]^2=(55)^2=3025\)
Kerala CEE-2016
Sequence and Series
118890
The value of \(\frac{1}{\sqrt{10}-\sqrt{9}}-\frac{1}{\sqrt{11}-\sqrt{10}}+\frac{1}{\sqrt{12}-\sqrt{11}}-\ldots .-\frac{1}{\sqrt{121}-\sqrt{120}}\) is equal to
118891
The sum of the series \(\frac{1^2}{1.2}+\frac{1^2+2^2}{2.3}+\frac{1^2+2^2+3^2}{3.4}+\ldots\). upto 20 terms is
1 \(\frac{205}{3}\)
2 \(\frac{200}{3}\)
3 \(\frac{220}{3}\)
4 \(\frac{210}{3}\)
Explanation:
C Let, \(S=\frac{1^2}{1.2}+\frac{1^2+2^2}{2.3}+\frac{1^2+2^2+3^2}{3.4}+\ldots\) upto 20 terms Let \(t_n\) be \(n^{\text {th }}\) term of series. Then, \(\mathrm{t}_{\mathrm{n}}=\frac{1^2+2^2+3^2+\ldots . .+\mathrm{n}^2}{\mathrm{n}(\mathrm{n}+1)}=\frac{2 \mathrm{n}+1}{6}\) Taking summation on both sides, we get- \(\sum_{\mathrm{n}=1}^{20} \mathrm{t}_{\mathrm{n}} =\frac{2}{6} \sum_{\mathrm{n}=1}^{20} \mathrm{n}+\frac{1}{6} \sum_{\mathrm{n}=1}^{20} 1\) \(=\frac{1}{3} \times\left(\frac{20(20+1)}{2}\right)+\frac{1}{6} \times 20\) \(\sum_{\mathrm{n}=1}^{20} \mathrm{t}_{\mathrm{n}} =\frac{1}{3} \times\left(\frac{20 \times 21}{2}\right)+\left(\frac{10}{3}\right)=\frac{210}{3}+\frac{10}{3}=\frac{220}{3}\)
COMEDK-2014
Sequence and Series
118892
\(11^3+12^3+13^3+\ldots \ldots \ldots \ldots . .+20^3\) is
1 an even integer
2 an odd integer divisible by 5
3 multiple of 10
4 an odd integer but not a multiple of 5
Explanation:
B Given that,\(11^3+12^3+13^3+\ldots .+20^3\) \(=\left(1^3+2^3+\ldots .+10^3+11^3+12^3+\ldots .+20^3\right)-\left(1^3+2^3+\ldots .+10^3\right)\) \(=\sum(20)^3-\sum(10)^3\) \(=\left(\frac{20(20+1)}{2}\right)^2-\left(\frac{10(10+1)}{2}\right)^2\) \(=(10 \times 21)^2-(5 \times 11)^2=(210-55)(210+55)\) \(=155 \times 265=41075=5 \times 8215\)
118888
In a bank, the principal increases continuously at the rate of \(6 \%\) per year. Then the time required to doubling \(₹ \mathbf{6 0 0 0}\) rupees is........ (in year)
1 \(\frac{50}{3} \log 2\)
2 \(\frac{30}{3} \log 6\)
3 \(\frac{50}{3} \log 3\)
4 \(\frac{50}{3} \log 12\)
Explanation:
A According to question, let \(\mathrm{P}\) and \(\mathrm{t}\) be the principle and time. \(P=6000\) rupees getting double in time ' \(t\) ' with rate of \(6 \%\) per year, So \(\frac{\mathrm{dP}}{\mathrm{dt}}=\frac{6}{100} \mathrm{P}\) \(\int_{6000}^{12000} \frac{\mathrm{dP}}{\mathrm{P}}=\frac{3}{50} \int_0^{\mathrm{t}} \mathrm{dt}\) \({[\log \mathrm{P}]_{6000}^{12000}=\frac{3}{50} \mathrm{t}}\) \(\log 12000-\log 6000=\frac{3}{50} \mathrm{t}\) \(\log 2=\frac{3}{50} \mathrm{t}\)Time, \(t=\frac{50}{3} \log 2\)
AP EAMCET-21.09.2020
Sequence and Series
118889
Sum of the series \(1(1)+2(1+3)+3(1+3+5)\) \(+4(1+3+5+7)+\ldots .+10(1+3+5+7\) \(+\ldots .+19)\) is equal to
1 385
2 1025
3 1125
4 2025
5 3025
Explanation:
E Here, we have \(1(1)+2(1+3)+3(1+3+5)+4(1+3+5+7)+\ldots .\) \(+10(1+3+5+7+\ldots .+19)\) \(=1 \times 1+2 \times 2^2+3 \times 3^2+4 \times 4^2+\ldots . .+10 \times 10^2\) \(=\left[\frac{10(10+1)}{2}\right]^2 \quad\left[\begin{array}{l}\because \text { Sum of cubes of } \mathrm{n} \text { natural } \\ \text { numbers }=\left[\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right]^2\end{array}\right]\) \(=\left[\frac{10 \times 11}{2}\right]^2=(55)^2=3025\)
Kerala CEE-2016
Sequence and Series
118890
The value of \(\frac{1}{\sqrt{10}-\sqrt{9}}-\frac{1}{\sqrt{11}-\sqrt{10}}+\frac{1}{\sqrt{12}-\sqrt{11}}-\ldots .-\frac{1}{\sqrt{121}-\sqrt{120}}\) is equal to
118891
The sum of the series \(\frac{1^2}{1.2}+\frac{1^2+2^2}{2.3}+\frac{1^2+2^2+3^2}{3.4}+\ldots\). upto 20 terms is
1 \(\frac{205}{3}\)
2 \(\frac{200}{3}\)
3 \(\frac{220}{3}\)
4 \(\frac{210}{3}\)
Explanation:
C Let, \(S=\frac{1^2}{1.2}+\frac{1^2+2^2}{2.3}+\frac{1^2+2^2+3^2}{3.4}+\ldots\) upto 20 terms Let \(t_n\) be \(n^{\text {th }}\) term of series. Then, \(\mathrm{t}_{\mathrm{n}}=\frac{1^2+2^2+3^2+\ldots . .+\mathrm{n}^2}{\mathrm{n}(\mathrm{n}+1)}=\frac{2 \mathrm{n}+1}{6}\) Taking summation on both sides, we get- \(\sum_{\mathrm{n}=1}^{20} \mathrm{t}_{\mathrm{n}} =\frac{2}{6} \sum_{\mathrm{n}=1}^{20} \mathrm{n}+\frac{1}{6} \sum_{\mathrm{n}=1}^{20} 1\) \(=\frac{1}{3} \times\left(\frac{20(20+1)}{2}\right)+\frac{1}{6} \times 20\) \(\sum_{\mathrm{n}=1}^{20} \mathrm{t}_{\mathrm{n}} =\frac{1}{3} \times\left(\frac{20 \times 21}{2}\right)+\left(\frac{10}{3}\right)=\frac{210}{3}+\frac{10}{3}=\frac{220}{3}\)
COMEDK-2014
Sequence and Series
118892
\(11^3+12^3+13^3+\ldots \ldots \ldots \ldots . .+20^3\) is
1 an even integer
2 an odd integer divisible by 5
3 multiple of 10
4 an odd integer but not a multiple of 5
Explanation:
B Given that,\(11^3+12^3+13^3+\ldots .+20^3\) \(=\left(1^3+2^3+\ldots .+10^3+11^3+12^3+\ldots .+20^3\right)-\left(1^3+2^3+\ldots .+10^3\right)\) \(=\sum(20)^3-\sum(10)^3\) \(=\left(\frac{20(20+1)}{2}\right)^2-\left(\frac{10(10+1)}{2}\right)^2\) \(=(10 \times 21)^2-(5 \times 11)^2=(210-55)(210+55)\) \(=155 \times 265=41075=5 \times 8215\)
118888
In a bank, the principal increases continuously at the rate of \(6 \%\) per year. Then the time required to doubling \(₹ \mathbf{6 0 0 0}\) rupees is........ (in year)
1 \(\frac{50}{3} \log 2\)
2 \(\frac{30}{3} \log 6\)
3 \(\frac{50}{3} \log 3\)
4 \(\frac{50}{3} \log 12\)
Explanation:
A According to question, let \(\mathrm{P}\) and \(\mathrm{t}\) be the principle and time. \(P=6000\) rupees getting double in time ' \(t\) ' with rate of \(6 \%\) per year, So \(\frac{\mathrm{dP}}{\mathrm{dt}}=\frac{6}{100} \mathrm{P}\) \(\int_{6000}^{12000} \frac{\mathrm{dP}}{\mathrm{P}}=\frac{3}{50} \int_0^{\mathrm{t}} \mathrm{dt}\) \({[\log \mathrm{P}]_{6000}^{12000}=\frac{3}{50} \mathrm{t}}\) \(\log 12000-\log 6000=\frac{3}{50} \mathrm{t}\) \(\log 2=\frac{3}{50} \mathrm{t}\)Time, \(t=\frac{50}{3} \log 2\)
AP EAMCET-21.09.2020
Sequence and Series
118889
Sum of the series \(1(1)+2(1+3)+3(1+3+5)\) \(+4(1+3+5+7)+\ldots .+10(1+3+5+7\) \(+\ldots .+19)\) is equal to
1 385
2 1025
3 1125
4 2025
5 3025
Explanation:
E Here, we have \(1(1)+2(1+3)+3(1+3+5)+4(1+3+5+7)+\ldots .\) \(+10(1+3+5+7+\ldots .+19)\) \(=1 \times 1+2 \times 2^2+3 \times 3^2+4 \times 4^2+\ldots . .+10 \times 10^2\) \(=\left[\frac{10(10+1)}{2}\right]^2 \quad\left[\begin{array}{l}\because \text { Sum of cubes of } \mathrm{n} \text { natural } \\ \text { numbers }=\left[\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right]^2\end{array}\right]\) \(=\left[\frac{10 \times 11}{2}\right]^2=(55)^2=3025\)
Kerala CEE-2016
Sequence and Series
118890
The value of \(\frac{1}{\sqrt{10}-\sqrt{9}}-\frac{1}{\sqrt{11}-\sqrt{10}}+\frac{1}{\sqrt{12}-\sqrt{11}}-\ldots .-\frac{1}{\sqrt{121}-\sqrt{120}}\) is equal to
118891
The sum of the series \(\frac{1^2}{1.2}+\frac{1^2+2^2}{2.3}+\frac{1^2+2^2+3^2}{3.4}+\ldots\). upto 20 terms is
1 \(\frac{205}{3}\)
2 \(\frac{200}{3}\)
3 \(\frac{220}{3}\)
4 \(\frac{210}{3}\)
Explanation:
C Let, \(S=\frac{1^2}{1.2}+\frac{1^2+2^2}{2.3}+\frac{1^2+2^2+3^2}{3.4}+\ldots\) upto 20 terms Let \(t_n\) be \(n^{\text {th }}\) term of series. Then, \(\mathrm{t}_{\mathrm{n}}=\frac{1^2+2^2+3^2+\ldots . .+\mathrm{n}^2}{\mathrm{n}(\mathrm{n}+1)}=\frac{2 \mathrm{n}+1}{6}\) Taking summation on both sides, we get- \(\sum_{\mathrm{n}=1}^{20} \mathrm{t}_{\mathrm{n}} =\frac{2}{6} \sum_{\mathrm{n}=1}^{20} \mathrm{n}+\frac{1}{6} \sum_{\mathrm{n}=1}^{20} 1\) \(=\frac{1}{3} \times\left(\frac{20(20+1)}{2}\right)+\frac{1}{6} \times 20\) \(\sum_{\mathrm{n}=1}^{20} \mathrm{t}_{\mathrm{n}} =\frac{1}{3} \times\left(\frac{20 \times 21}{2}\right)+\left(\frac{10}{3}\right)=\frac{210}{3}+\frac{10}{3}=\frac{220}{3}\)
COMEDK-2014
Sequence and Series
118892
\(11^3+12^3+13^3+\ldots \ldots \ldots \ldots . .+20^3\) is
1 an even integer
2 an odd integer divisible by 5
3 multiple of 10
4 an odd integer but not a multiple of 5
Explanation:
B Given that,\(11^3+12^3+13^3+\ldots .+20^3\) \(=\left(1^3+2^3+\ldots .+10^3+11^3+12^3+\ldots .+20^3\right)-\left(1^3+2^3+\ldots .+10^3\right)\) \(=\sum(20)^3-\sum(10)^3\) \(=\left(\frac{20(20+1)}{2}\right)^2-\left(\frac{10(10+1)}{2}\right)^2\) \(=(10 \times 21)^2-(5 \times 11)^2=(210-55)(210+55)\) \(=155 \times 265=41075=5 \times 8215\)
