119380
If the \(9^{\text {th }}\) and \(10^{\text {th }}\) terms are the numerically greatest terms in the expansion of \((5 x-6 y)^n\) when \(x=2 / 5\) and \(y=1 / 2\), then the absolute value of the middle terms of that expansion is
1 \({ }^{14} \mathrm{C}_8 6^7\)
2 \({ }^{14} \mathrm{C}_7 6^7\)
3 \({ }^{15} \mathrm{C}_7 6^7\)
4 \({ }^{15} \mathrm{C}_8 6\)
Explanation:
B Given, \(9^{\text {th }}\) and \(10^{\text {th }}\) terms are numerically greatest terms in the expansion of \((5 x-6 y)^n\). \(9 \leq \frac{(\mathrm{n}+1)\left(\frac{6 \mathrm{y}}{5 \mathrm{x}}\right)}{1+\left(\frac{6 \mathrm{y}}{5 \mathrm{x}}\right)}\) Here, \(\mathrm{y}=\frac{1}{2}\) and \(\mathrm{x}=\frac{2}{5}\) \(\because \quad \frac{\mathrm{y}}{\mathrm{x}}=\frac{5}{4} \Rightarrow \frac{6 \mathrm{y}}{5 \mathrm{x}}=\frac{3}{2}\) \(9 \leq \frac{(\mathrm{n}+1)\left(\frac{3}{2}\right)}{1+\frac{3}{2}}\) \(9 \leq \frac{3(\mathrm{n}+1)}{5}\) \(\mathrm{n}+1 \geq 15\) \(\mathrm{n} \geq 14\) \(\mathrm{n}=14\) \(\because \quad \mathrm{n}=14\) Middle term of \((5 \mathrm{x}-6 \mathrm{y})^{14}\) \(=\left|{ }^{14} \mathrm{C}_7(5 \mathrm{x})^7(-6 \mathrm{y})^7\right|\) \(=\left|{ }^{14} \mathrm{C}_7\left(5 \times \frac{2}{5}\right)^7\left(-6 \times \frac{1}{2}\right)^7\right|={ }^{14} \mathrm{C}_7 \times 6^7\)
TS EAMCET-10.09.2020
Binomial Theorem and its Simple Application
119382
Let \(x \in R\) be so small that the powers of \(x\) beyond two are insignificant and negligibly small. For such \(x\), if \((1-x)^3,(2+x)^6\) is approximated by \(\mathbf{a}+\mathbf{b x}+\mathbf{c x}^2\), then \(\mathbf{a}+\mathbf{b}+\mathbf{c}=\)
119383
If \(x^{22}\) is in the \((r+1)^{\text {th }}\) term of the binomial expansion of \(\left(3 x^3-x^2\right)^9\), then the value of \(r\) is equal to
119385
The remainder when \(2^{2000}\) is divided by 17 is
1 1
2 2
3 8
4 12
5 4
Explanation:
A Given, \(2^{2000}\) It can be written, \(\left(2^4\right)^{500}\) \(=(16)^{500}\) \(=(17-1)^{500}\) \(=(-1)^{500}=(1)\) If \(2^{2000}\) is divided by 17 Hence, reminder is 1
119380
If the \(9^{\text {th }}\) and \(10^{\text {th }}\) terms are the numerically greatest terms in the expansion of \((5 x-6 y)^n\) when \(x=2 / 5\) and \(y=1 / 2\), then the absolute value of the middle terms of that expansion is
1 \({ }^{14} \mathrm{C}_8 6^7\)
2 \({ }^{14} \mathrm{C}_7 6^7\)
3 \({ }^{15} \mathrm{C}_7 6^7\)
4 \({ }^{15} \mathrm{C}_8 6\)
Explanation:
B Given, \(9^{\text {th }}\) and \(10^{\text {th }}\) terms are numerically greatest terms in the expansion of \((5 x-6 y)^n\). \(9 \leq \frac{(\mathrm{n}+1)\left(\frac{6 \mathrm{y}}{5 \mathrm{x}}\right)}{1+\left(\frac{6 \mathrm{y}}{5 \mathrm{x}}\right)}\) Here, \(\mathrm{y}=\frac{1}{2}\) and \(\mathrm{x}=\frac{2}{5}\) \(\because \quad \frac{\mathrm{y}}{\mathrm{x}}=\frac{5}{4} \Rightarrow \frac{6 \mathrm{y}}{5 \mathrm{x}}=\frac{3}{2}\) \(9 \leq \frac{(\mathrm{n}+1)\left(\frac{3}{2}\right)}{1+\frac{3}{2}}\) \(9 \leq \frac{3(\mathrm{n}+1)}{5}\) \(\mathrm{n}+1 \geq 15\) \(\mathrm{n} \geq 14\) \(\mathrm{n}=14\) \(\because \quad \mathrm{n}=14\) Middle term of \((5 \mathrm{x}-6 \mathrm{y})^{14}\) \(=\left|{ }^{14} \mathrm{C}_7(5 \mathrm{x})^7(-6 \mathrm{y})^7\right|\) \(=\left|{ }^{14} \mathrm{C}_7\left(5 \times \frac{2}{5}\right)^7\left(-6 \times \frac{1}{2}\right)^7\right|={ }^{14} \mathrm{C}_7 \times 6^7\)
TS EAMCET-10.09.2020
Binomial Theorem and its Simple Application
119382
Let \(x \in R\) be so small that the powers of \(x\) beyond two are insignificant and negligibly small. For such \(x\), if \((1-x)^3,(2+x)^6\) is approximated by \(\mathbf{a}+\mathbf{b x}+\mathbf{c x}^2\), then \(\mathbf{a}+\mathbf{b}+\mathbf{c}=\)
119383
If \(x^{22}\) is in the \((r+1)^{\text {th }}\) term of the binomial expansion of \(\left(3 x^3-x^2\right)^9\), then the value of \(r\) is equal to
119385
The remainder when \(2^{2000}\) is divided by 17 is
1 1
2 2
3 8
4 12
5 4
Explanation:
A Given, \(2^{2000}\) It can be written, \(\left(2^4\right)^{500}\) \(=(16)^{500}\) \(=(17-1)^{500}\) \(=(-1)^{500}=(1)\) If \(2^{2000}\) is divided by 17 Hence, reminder is 1
119380
If the \(9^{\text {th }}\) and \(10^{\text {th }}\) terms are the numerically greatest terms in the expansion of \((5 x-6 y)^n\) when \(x=2 / 5\) and \(y=1 / 2\), then the absolute value of the middle terms of that expansion is
1 \({ }^{14} \mathrm{C}_8 6^7\)
2 \({ }^{14} \mathrm{C}_7 6^7\)
3 \({ }^{15} \mathrm{C}_7 6^7\)
4 \({ }^{15} \mathrm{C}_8 6\)
Explanation:
B Given, \(9^{\text {th }}\) and \(10^{\text {th }}\) terms are numerically greatest terms in the expansion of \((5 x-6 y)^n\). \(9 \leq \frac{(\mathrm{n}+1)\left(\frac{6 \mathrm{y}}{5 \mathrm{x}}\right)}{1+\left(\frac{6 \mathrm{y}}{5 \mathrm{x}}\right)}\) Here, \(\mathrm{y}=\frac{1}{2}\) and \(\mathrm{x}=\frac{2}{5}\) \(\because \quad \frac{\mathrm{y}}{\mathrm{x}}=\frac{5}{4} \Rightarrow \frac{6 \mathrm{y}}{5 \mathrm{x}}=\frac{3}{2}\) \(9 \leq \frac{(\mathrm{n}+1)\left(\frac{3}{2}\right)}{1+\frac{3}{2}}\) \(9 \leq \frac{3(\mathrm{n}+1)}{5}\) \(\mathrm{n}+1 \geq 15\) \(\mathrm{n} \geq 14\) \(\mathrm{n}=14\) \(\because \quad \mathrm{n}=14\) Middle term of \((5 \mathrm{x}-6 \mathrm{y})^{14}\) \(=\left|{ }^{14} \mathrm{C}_7(5 \mathrm{x})^7(-6 \mathrm{y})^7\right|\) \(=\left|{ }^{14} \mathrm{C}_7\left(5 \times \frac{2}{5}\right)^7\left(-6 \times \frac{1}{2}\right)^7\right|={ }^{14} \mathrm{C}_7 \times 6^7\)
TS EAMCET-10.09.2020
Binomial Theorem and its Simple Application
119382
Let \(x \in R\) be so small that the powers of \(x\) beyond two are insignificant and negligibly small. For such \(x\), if \((1-x)^3,(2+x)^6\) is approximated by \(\mathbf{a}+\mathbf{b x}+\mathbf{c x}^2\), then \(\mathbf{a}+\mathbf{b}+\mathbf{c}=\)
119383
If \(x^{22}\) is in the \((r+1)^{\text {th }}\) term of the binomial expansion of \(\left(3 x^3-x^2\right)^9\), then the value of \(r\) is equal to
119385
The remainder when \(2^{2000}\) is divided by 17 is
1 1
2 2
3 8
4 12
5 4
Explanation:
A Given, \(2^{2000}\) It can be written, \(\left(2^4\right)^{500}\) \(=(16)^{500}\) \(=(17-1)^{500}\) \(=(-1)^{500}=(1)\) If \(2^{2000}\) is divided by 17 Hence, reminder is 1
119380
If the \(9^{\text {th }}\) and \(10^{\text {th }}\) terms are the numerically greatest terms in the expansion of \((5 x-6 y)^n\) when \(x=2 / 5\) and \(y=1 / 2\), then the absolute value of the middle terms of that expansion is
1 \({ }^{14} \mathrm{C}_8 6^7\)
2 \({ }^{14} \mathrm{C}_7 6^7\)
3 \({ }^{15} \mathrm{C}_7 6^7\)
4 \({ }^{15} \mathrm{C}_8 6\)
Explanation:
B Given, \(9^{\text {th }}\) and \(10^{\text {th }}\) terms are numerically greatest terms in the expansion of \((5 x-6 y)^n\). \(9 \leq \frac{(\mathrm{n}+1)\left(\frac{6 \mathrm{y}}{5 \mathrm{x}}\right)}{1+\left(\frac{6 \mathrm{y}}{5 \mathrm{x}}\right)}\) Here, \(\mathrm{y}=\frac{1}{2}\) and \(\mathrm{x}=\frac{2}{5}\) \(\because \quad \frac{\mathrm{y}}{\mathrm{x}}=\frac{5}{4} \Rightarrow \frac{6 \mathrm{y}}{5 \mathrm{x}}=\frac{3}{2}\) \(9 \leq \frac{(\mathrm{n}+1)\left(\frac{3}{2}\right)}{1+\frac{3}{2}}\) \(9 \leq \frac{3(\mathrm{n}+1)}{5}\) \(\mathrm{n}+1 \geq 15\) \(\mathrm{n} \geq 14\) \(\mathrm{n}=14\) \(\because \quad \mathrm{n}=14\) Middle term of \((5 \mathrm{x}-6 \mathrm{y})^{14}\) \(=\left|{ }^{14} \mathrm{C}_7(5 \mathrm{x})^7(-6 \mathrm{y})^7\right|\) \(=\left|{ }^{14} \mathrm{C}_7\left(5 \times \frac{2}{5}\right)^7\left(-6 \times \frac{1}{2}\right)^7\right|={ }^{14} \mathrm{C}_7 \times 6^7\)
TS EAMCET-10.09.2020
Binomial Theorem and its Simple Application
119382
Let \(x \in R\) be so small that the powers of \(x\) beyond two are insignificant and negligibly small. For such \(x\), if \((1-x)^3,(2+x)^6\) is approximated by \(\mathbf{a}+\mathbf{b x}+\mathbf{c x}^2\), then \(\mathbf{a}+\mathbf{b}+\mathbf{c}=\)
119383
If \(x^{22}\) is in the \((r+1)^{\text {th }}\) term of the binomial expansion of \(\left(3 x^3-x^2\right)^9\), then the value of \(r\) is equal to
119385
The remainder when \(2^{2000}\) is divided by 17 is
1 1
2 2
3 8
4 12
5 4
Explanation:
A Given, \(2^{2000}\) It can be written, \(\left(2^4\right)^{500}\) \(=(16)^{500}\) \(=(17-1)^{500}\) \(=(-1)^{500}=(1)\) If \(2^{2000}\) is divided by 17 Hence, reminder is 1
119380
If the \(9^{\text {th }}\) and \(10^{\text {th }}\) terms are the numerically greatest terms in the expansion of \((5 x-6 y)^n\) when \(x=2 / 5\) and \(y=1 / 2\), then the absolute value of the middle terms of that expansion is
1 \({ }^{14} \mathrm{C}_8 6^7\)
2 \({ }^{14} \mathrm{C}_7 6^7\)
3 \({ }^{15} \mathrm{C}_7 6^7\)
4 \({ }^{15} \mathrm{C}_8 6\)
Explanation:
B Given, \(9^{\text {th }}\) and \(10^{\text {th }}\) terms are numerically greatest terms in the expansion of \((5 x-6 y)^n\). \(9 \leq \frac{(\mathrm{n}+1)\left(\frac{6 \mathrm{y}}{5 \mathrm{x}}\right)}{1+\left(\frac{6 \mathrm{y}}{5 \mathrm{x}}\right)}\) Here, \(\mathrm{y}=\frac{1}{2}\) and \(\mathrm{x}=\frac{2}{5}\) \(\because \quad \frac{\mathrm{y}}{\mathrm{x}}=\frac{5}{4} \Rightarrow \frac{6 \mathrm{y}}{5 \mathrm{x}}=\frac{3}{2}\) \(9 \leq \frac{(\mathrm{n}+1)\left(\frac{3}{2}\right)}{1+\frac{3}{2}}\) \(9 \leq \frac{3(\mathrm{n}+1)}{5}\) \(\mathrm{n}+1 \geq 15\) \(\mathrm{n} \geq 14\) \(\mathrm{n}=14\) \(\because \quad \mathrm{n}=14\) Middle term of \((5 \mathrm{x}-6 \mathrm{y})^{14}\) \(=\left|{ }^{14} \mathrm{C}_7(5 \mathrm{x})^7(-6 \mathrm{y})^7\right|\) \(=\left|{ }^{14} \mathrm{C}_7\left(5 \times \frac{2}{5}\right)^7\left(-6 \times \frac{1}{2}\right)^7\right|={ }^{14} \mathrm{C}_7 \times 6^7\)
TS EAMCET-10.09.2020
Binomial Theorem and its Simple Application
119382
Let \(x \in R\) be so small that the powers of \(x\) beyond two are insignificant and negligibly small. For such \(x\), if \((1-x)^3,(2+x)^6\) is approximated by \(\mathbf{a}+\mathbf{b x}+\mathbf{c x}^2\), then \(\mathbf{a}+\mathbf{b}+\mathbf{c}=\)
119383
If \(x^{22}\) is in the \((r+1)^{\text {th }}\) term of the binomial expansion of \(\left(3 x^3-x^2\right)^9\), then the value of \(r\) is equal to
119385
The remainder when \(2^{2000}\) is divided by 17 is
1 1
2 2
3 8
4 12
5 4
Explanation:
A Given, \(2^{2000}\) It can be written, \(\left(2^4\right)^{500}\) \(=(16)^{500}\) \(=(17-1)^{500}\) \(=(-1)^{500}=(1)\) If \(2^{2000}\) is divided by 17 Hence, reminder is 1
