118672
The AM of 9 terms is 15 . If one more term is added to this series, then the AM becomes 16 the value so the added term is
1 30
2 27
3 25
4 23
5 20
Explanation:
C Given, A.M of 9 terms \(=15\) \(\frac{a_1+\ldots .+a_9}{9}=15\) \(a_1+a_2+\ldots .+a_9=135\) \(\mathrm{AM}\) of 10 terms \(=16\) \(\frac{a_1+a_2+\ldots .+a_{10}}{10}=16\) \(\left(a_1+a_2+\ldots .+a_9\right)+a_{10}=160\) On putting the value equation (i) in equation (ii), \(135+a_{10}=160\) \(a_{10}=25\)So, the added term is 25 .
Kerala CEE-2011
Sequence and Series
118674
The arithmetic mean of 7 consecutive integers starting with ' \(a\) ' is \(\mathrm{m}\). Then, the arithmetic mean of 11 consecutive integers starting with 'a +2 ' is
118675
If \(a, b, c\) are in GP and \(x, y\) are arithmetic mean of \(a, b\) and \(b, c\) respectively, then \(\frac{1}{x}+\frac{1}{y}\) is equal to
1 \(\frac{2}{\mathrm{~b}}\)
2 \(\frac{3}{\mathrm{~b}}\)
3 \(\frac{\mathrm{b}}{3}\)
4 \(\frac{\mathrm{b}}{2}\)
5 \(\frac{1}{\mathrm{~b}}\)
Explanation:
A Given, If \(a, b, c\) are in G.P. Then, \(\frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{c}}{\mathrm{b}}\) \(\mathrm{b}^2=\mathrm{ac}\) If \(x, y\) are arithmetic mean of \(a, b\) and \(b, c\) respectively. Then, \(\quad x=\frac{a+b}{2}\) and \(y=\frac{b+c}{2}\) \(\therefore \quad \frac{1}{x}+\frac{1}{y} =\frac{2}{a+b}+\frac{2}{b+c}\) \(\frac{1}{x}+\frac{1}{y} =\frac{2(2 b+a+c)}{(a+b)(b+c)}\) \(\frac{1}{x}+\frac{1}{y} =\frac{2(2 b+a+c)}{a b+b^2+b c+a c}\) On putting the value ac in above equation. \(\frac{1}{x}+\frac{1}{y}=\frac{2(2 b+a+c)}{a b+b^2+b c+b^2}\) \(\frac{1}{x}+\frac{1}{y}=\frac{2(2 b+a+c)}{a b+2 b^2+b c}\) \(\frac{1}{x}+\frac{1}{y}=\frac{2(2 b+a+c)}{b(2 b+a+c)}\) \(\frac{1}{x}+\frac{1}{y}=\frac{2}{b}\)
Kerala CEE-2009
Sequence and Series
118676
If \(A, G, H\) are respectively the A.M., G.M., H.M. of three numbers \(\alpha, \beta, \gamma\), the equation whose roots are \(\alpha, \beta, \gamma\), is
118672
The AM of 9 terms is 15 . If one more term is added to this series, then the AM becomes 16 the value so the added term is
1 30
2 27
3 25
4 23
5 20
Explanation:
C Given, A.M of 9 terms \(=15\) \(\frac{a_1+\ldots .+a_9}{9}=15\) \(a_1+a_2+\ldots .+a_9=135\) \(\mathrm{AM}\) of 10 terms \(=16\) \(\frac{a_1+a_2+\ldots .+a_{10}}{10}=16\) \(\left(a_1+a_2+\ldots .+a_9\right)+a_{10}=160\) On putting the value equation (i) in equation (ii), \(135+a_{10}=160\) \(a_{10}=25\)So, the added term is 25 .
Kerala CEE-2011
Sequence and Series
118674
The arithmetic mean of 7 consecutive integers starting with ' \(a\) ' is \(\mathrm{m}\). Then, the arithmetic mean of 11 consecutive integers starting with 'a +2 ' is
118675
If \(a, b, c\) are in GP and \(x, y\) are arithmetic mean of \(a, b\) and \(b, c\) respectively, then \(\frac{1}{x}+\frac{1}{y}\) is equal to
1 \(\frac{2}{\mathrm{~b}}\)
2 \(\frac{3}{\mathrm{~b}}\)
3 \(\frac{\mathrm{b}}{3}\)
4 \(\frac{\mathrm{b}}{2}\)
5 \(\frac{1}{\mathrm{~b}}\)
Explanation:
A Given, If \(a, b, c\) are in G.P. Then, \(\frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{c}}{\mathrm{b}}\) \(\mathrm{b}^2=\mathrm{ac}\) If \(x, y\) are arithmetic mean of \(a, b\) and \(b, c\) respectively. Then, \(\quad x=\frac{a+b}{2}\) and \(y=\frac{b+c}{2}\) \(\therefore \quad \frac{1}{x}+\frac{1}{y} =\frac{2}{a+b}+\frac{2}{b+c}\) \(\frac{1}{x}+\frac{1}{y} =\frac{2(2 b+a+c)}{(a+b)(b+c)}\) \(\frac{1}{x}+\frac{1}{y} =\frac{2(2 b+a+c)}{a b+b^2+b c+a c}\) On putting the value ac in above equation. \(\frac{1}{x}+\frac{1}{y}=\frac{2(2 b+a+c)}{a b+b^2+b c+b^2}\) \(\frac{1}{x}+\frac{1}{y}=\frac{2(2 b+a+c)}{a b+2 b^2+b c}\) \(\frac{1}{x}+\frac{1}{y}=\frac{2(2 b+a+c)}{b(2 b+a+c)}\) \(\frac{1}{x}+\frac{1}{y}=\frac{2}{b}\)
Kerala CEE-2009
Sequence and Series
118676
If \(A, G, H\) are respectively the A.M., G.M., H.M. of three numbers \(\alpha, \beta, \gamma\), the equation whose roots are \(\alpha, \beta, \gamma\), is
118672
The AM of 9 terms is 15 . If one more term is added to this series, then the AM becomes 16 the value so the added term is
1 30
2 27
3 25
4 23
5 20
Explanation:
C Given, A.M of 9 terms \(=15\) \(\frac{a_1+\ldots .+a_9}{9}=15\) \(a_1+a_2+\ldots .+a_9=135\) \(\mathrm{AM}\) of 10 terms \(=16\) \(\frac{a_1+a_2+\ldots .+a_{10}}{10}=16\) \(\left(a_1+a_2+\ldots .+a_9\right)+a_{10}=160\) On putting the value equation (i) in equation (ii), \(135+a_{10}=160\) \(a_{10}=25\)So, the added term is 25 .
Kerala CEE-2011
Sequence and Series
118674
The arithmetic mean of 7 consecutive integers starting with ' \(a\) ' is \(\mathrm{m}\). Then, the arithmetic mean of 11 consecutive integers starting with 'a +2 ' is
118675
If \(a, b, c\) are in GP and \(x, y\) are arithmetic mean of \(a, b\) and \(b, c\) respectively, then \(\frac{1}{x}+\frac{1}{y}\) is equal to
1 \(\frac{2}{\mathrm{~b}}\)
2 \(\frac{3}{\mathrm{~b}}\)
3 \(\frac{\mathrm{b}}{3}\)
4 \(\frac{\mathrm{b}}{2}\)
5 \(\frac{1}{\mathrm{~b}}\)
Explanation:
A Given, If \(a, b, c\) are in G.P. Then, \(\frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{c}}{\mathrm{b}}\) \(\mathrm{b}^2=\mathrm{ac}\) If \(x, y\) are arithmetic mean of \(a, b\) and \(b, c\) respectively. Then, \(\quad x=\frac{a+b}{2}\) and \(y=\frac{b+c}{2}\) \(\therefore \quad \frac{1}{x}+\frac{1}{y} =\frac{2}{a+b}+\frac{2}{b+c}\) \(\frac{1}{x}+\frac{1}{y} =\frac{2(2 b+a+c)}{(a+b)(b+c)}\) \(\frac{1}{x}+\frac{1}{y} =\frac{2(2 b+a+c)}{a b+b^2+b c+a c}\) On putting the value ac in above equation. \(\frac{1}{x}+\frac{1}{y}=\frac{2(2 b+a+c)}{a b+b^2+b c+b^2}\) \(\frac{1}{x}+\frac{1}{y}=\frac{2(2 b+a+c)}{a b+2 b^2+b c}\) \(\frac{1}{x}+\frac{1}{y}=\frac{2(2 b+a+c)}{b(2 b+a+c)}\) \(\frac{1}{x}+\frac{1}{y}=\frac{2}{b}\)
Kerala CEE-2009
Sequence and Series
118676
If \(A, G, H\) are respectively the A.M., G.M., H.M. of three numbers \(\alpha, \beta, \gamma\), the equation whose roots are \(\alpha, \beta, \gamma\), is
NEET Test Series from KOTA - 10 Papers In MS WORD
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Sequence and Series
118672
The AM of 9 terms is 15 . If one more term is added to this series, then the AM becomes 16 the value so the added term is
1 30
2 27
3 25
4 23
5 20
Explanation:
C Given, A.M of 9 terms \(=15\) \(\frac{a_1+\ldots .+a_9}{9}=15\) \(a_1+a_2+\ldots .+a_9=135\) \(\mathrm{AM}\) of 10 terms \(=16\) \(\frac{a_1+a_2+\ldots .+a_{10}}{10}=16\) \(\left(a_1+a_2+\ldots .+a_9\right)+a_{10}=160\) On putting the value equation (i) in equation (ii), \(135+a_{10}=160\) \(a_{10}=25\)So, the added term is 25 .
Kerala CEE-2011
Sequence and Series
118674
The arithmetic mean of 7 consecutive integers starting with ' \(a\) ' is \(\mathrm{m}\). Then, the arithmetic mean of 11 consecutive integers starting with 'a +2 ' is
118675
If \(a, b, c\) are in GP and \(x, y\) are arithmetic mean of \(a, b\) and \(b, c\) respectively, then \(\frac{1}{x}+\frac{1}{y}\) is equal to
1 \(\frac{2}{\mathrm{~b}}\)
2 \(\frac{3}{\mathrm{~b}}\)
3 \(\frac{\mathrm{b}}{3}\)
4 \(\frac{\mathrm{b}}{2}\)
5 \(\frac{1}{\mathrm{~b}}\)
Explanation:
A Given, If \(a, b, c\) are in G.P. Then, \(\frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{c}}{\mathrm{b}}\) \(\mathrm{b}^2=\mathrm{ac}\) If \(x, y\) are arithmetic mean of \(a, b\) and \(b, c\) respectively. Then, \(\quad x=\frac{a+b}{2}\) and \(y=\frac{b+c}{2}\) \(\therefore \quad \frac{1}{x}+\frac{1}{y} =\frac{2}{a+b}+\frac{2}{b+c}\) \(\frac{1}{x}+\frac{1}{y} =\frac{2(2 b+a+c)}{(a+b)(b+c)}\) \(\frac{1}{x}+\frac{1}{y} =\frac{2(2 b+a+c)}{a b+b^2+b c+a c}\) On putting the value ac in above equation. \(\frac{1}{x}+\frac{1}{y}=\frac{2(2 b+a+c)}{a b+b^2+b c+b^2}\) \(\frac{1}{x}+\frac{1}{y}=\frac{2(2 b+a+c)}{a b+2 b^2+b c}\) \(\frac{1}{x}+\frac{1}{y}=\frac{2(2 b+a+c)}{b(2 b+a+c)}\) \(\frac{1}{x}+\frac{1}{y}=\frac{2}{b}\)
Kerala CEE-2009
Sequence and Series
118676
If \(A, G, H\) are respectively the A.M., G.M., H.M. of three numbers \(\alpha, \beta, \gamma\), the equation whose roots are \(\alpha, \beta, \gamma\), is
