Since \(a,b,c\) are in AP, hence \(2b = a + c\).
Now \(\left( {{e^{\frac{1}{c}}}} \right)\left( {{e^{\frac{1}{a}}}} \right) = {e^{\frac{{a + c}}{{ac}}}} = {e^{\frac{{2b}}{{ac}}}} = {\left( {{e^{\frac{b}{{ac}}}}} \right)^2}\)
So the numbers \({e^{\frac{1}{c}}},\,{e^{\frac{b}{{ac}}}},\,{e^{\frac{1}{a}}}\) are in GP.

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The numbers \(\frac{1}{a},\,\frac{1}{{{H_1}}},........\frac{1}{{{H_n}}},\,\frac{1}{b}\) are in AP, hence
\(\frac{1}{{{H_1}}} - \frac{1}{a} = \frac{1}{b} - \frac{1}{{{H_n}}} = d\) and \(\frac{1}{b} = \frac{1}{a} + (n + 1)d\)
Now \(\frac{{{H_1} + a}}{{{H_1} - a}} + \frac{{{H_n} + b}}{{{H_n} - b}} = \frac{{\frac{1}{a} + \frac{1}{{{H_1}}}}}{{\frac{1}{a} - \frac{1}{{{H_1}}}}} + \frac{{\frac{1}{b} + \frac{1}{{{H_n}}}}}{{\frac{1}{b} - \frac{1}{{{H_n}}}}}\)
\( = \frac{{\frac{1}{a} + \frac{1}{a} + d}}{{ - d}} + \frac{{\frac{1}{b} + \frac{1}{b} - d}}{d}\)
\( = - \frac{2}{{ad}} - 1 + \frac{2}{{bd}} - 1 = \frac{2}{d}\left( {\frac{1}{b} - \frac{1}{a}} \right) - 2\)
\( = \frac{2}{d}\left[ {(n + 1)d} \right] - 2 = 2n\)

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Number of elements in \({A_1},\,{A_2},\,{A_3}\) are 1, 3, 5,…..
Hence total number of elements upto \({A_{19}}\) = 1 + 3 + 5 + 7 + …. + 19 terms = 19² = 361
(as sum of the first \(n\) odd numbers is \(n^2\))
So \({A_{20}}\)will contain 39 terms and will start with 362nd term of {3, 5, 7, 9, 11, ……}, that is 2×362 + 1 = 725.
Thus average of the series {725, 727, 729, ……..39 terms} will be 19th term of the series = 725 + 18×2 = 763

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From the given information, we have
\(2b = a + c,\,q = \frac{{2pr}}{{p + r}},\;{b^2}{q^2} = acpr\)
Now \(\frac{p}{r} + \frac{r}{p} = \frac{{{{(p + r)}^2} - 2pr}}{{pr}} = \frac{{{{(p + r)}^2}}}{{pr}} - 2\)
Putting the value of \(p + r = \frac{{2pr}}{q}\), we have
\(\frac{p}{r} + \frac{r}{p} = \frac{{{{(p + r)}^2}}}{{pr}} - 2 = \frac{{4pr}}{{{q^2}}} - 2\)
Put the value of \({q^2} = \frac{{acpr}}{{{b^2}}}\)
\(\frac{p}{r} + \frac{r}{p} = \frac{{4pr}}{{{q^2}}} - 2 = \frac{{4pr}}{{\frac{{acpr}}{{{b^2}}}}} - 2 = \frac{{4{b^2}}}{{ac}} - 2\)
\( = \frac{{{{(a + c)}^2}}}{{ac}} - 2 = \frac{{{a^2} + {c^2} + 2ac}}{{ac}} - 2 = \frac{a}{c} + \frac{c}{a}\)

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Mean of the squares of first \(n\) natural numbers = \(\frac{{n(n + 1)(2n + 1)}}{{6n}} = \frac{{(n + 1)(2n + 1)}}{6}\)
\( \Rightarrow \frac{{(n + 1)(2n + 1)}}{6} = 11\)
Or \(n = 5\)

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Mean of the first \(n\)natural numbers \( = \frac{{n(n + 1)}}{{2n}} = \frac{{n + 1}}{2}\)
Given that \(\frac{{n + 1}}{2} = \frac{{n + 7}}{3} \Rightarrow 3n + 3 = 2n + 14\)
Or \(n = 11\)

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Given that \(a,b,c\) are in HP, hence \(2b = \frac{{ac}}{{a + c}}\)
\( \Rightarrow \log (a - 2b + c) = \log \left( {a + c - \frac{{4ac}}{{a + c}}} \right)\)
\( = \log \left( {\frac{{{{(a + c)}^2} - 4ac}}{{a + c}}} \right) = \log \left( {\frac{{{{(c - a)}^2}}}{{a + c}}} \right) = 2\log (c - a) - \log (a + c)\)
\(\log (a + c) + \log (a - 2b + c) = 2\log (c - a)\)

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The required sum = \(\sum\limits_{n = 1}^{30} {{n^2}} - \sum\limits_{n = 1}^{10} {{n^2}} = \frac{{30 \times 31 \times 61}}{6} - \frac{{10 \times 11 \times 21}}{6}\)
\( = \frac{{30}}{6}\left( {31 \times 61 - 11 \times 7} \right) = 9070\)
The value of \({9^{\frac{1}{3}}}\,\,{9^{\frac{1}{9}}}\,\,\,{9^{\frac{1}{{27}}}}...........\infty \) is:

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The value is \({9^{\frac{1}{3} + \frac{1}{9} + \frac{1}{{27}} + .....\infty }} = {9^{\frac{{\frac{1}{3}}}{{\,\,1 - \frac{1}{3}\,\,}}}} = {9^{\frac{1}{2}}} = \sqrt 9 = 3\)

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The first number that is a multiple of 7 and more than 200 is 203. The last number is 399.
So the series is 203, 210, 217, ……. , 399
To find the number of terms in the series,
\(203 + (n - 1) \times 7 = 399\) or \(n = 29.\)
Sum of the series = \(\frac{{203 + 399}}{2} \times 29 = 8729\)

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The series is \( - {1^2} + {2^2} - {3^2} + {4^2} - ...... + {20^2}\)
Taking terms in pairs, we have
\(\underbrace { - {1^2} + {2^2}}_{}\underbrace { - {3^2} + {4^2}}_{} - ......\underbrace { - {{19}^2} + {{20}^2}}_{}\)
\({2^2} - {1^2} = 3,\,\,{4^2} - {3^2} = 7,\,\,{6^2} - {5^2} = 11,....\)
The required sum is \(3 + 7 + 11 + ..... + 39 = \left( {\frac{{3 + 39}}{2}} \right) \times 10 = 210\)

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Given that \(a,\,b,\,c\) are in GP, hence \({b^2} = ac\)
Multiplying by \({x^2}\) both sides,
\({x^2}{b^2} = (ax)(cx)\)
Taking log, both sides,
\({\log _x}({x^2}{b^2}) = {\log _x}(ax) + {\log _x}(cx)\)
\( \Rightarrow 2{\log _x}(bx) = {\log _x}(ax) + {\log _x}(cx)\)
\( \Rightarrow {\log _x}(bx) - {\log _x}(ax) = {\log _x}(cx) - {\log _x}(bx)\)
Hence \({\log _x}(ax),\,{\log _x}(bx),\,{\log _x}(cx)\,\) are in AP.
Or \({\log _{ax}}x,{\log _{bx}}x\) and \({\log _{cx}}x\) are in HP.

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Rationalize the denominators, we get the following series,
\(\frac{{\sqrt 2 - 1}}{1} + \frac{{\sqrt 3 - \sqrt 2 }}{1} + \frac{{\sqrt 4 - \sqrt 3 }}{1} + ...\frac{{\sqrt {81} - \sqrt {80} }}{1}\)
\( = \sqrt {81} - 1 = 8\)

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Given that \({a_1} = 0\), hence \({a_2} = d,\,{a_3} = 2d,\,{a_4} = 3d,...\;{\rm{etc}}\).
The value of \(\left( {\frac{{{a_3}}}{{{a_2}}} + \frac{{{a_4}}}{{{a_3}}} + .... + \frac{{{a_n}}}{{{a_{n - 1}}}}} \right) - {a_2}\left( {\frac{1}{{{a_2}}} + \frac{1}{{{a_3}}} + ..... + \frac{1}{{{a_{n - 2}}}}} \right)\)
\(\left( {\frac{{2d}}{d} + \frac{{3d}}{{2d}} + ..... + \frac{{(n - 1)d}}{{(n - 2)d}}} \right) - d\left( {\frac{1}{d} + \frac{1}{{2d}} + .. + \frac{1}{{n - 3}}} \right)\)
\( = \left( {\frac{2}{1} + \frac{3}{2} + .... + \frac{{n - 1}}{{n - 2}}} \right) - \left( {\frac{1}{1} + \frac{1}{2} + .... + \frac{1}{{n - 3}}} \right)\)
\( = \left( {\frac{2}{1} - \frac{1}{1}} \right) + \left( {\frac{3}{2} - \frac{1}{2}} \right) + ....\left( {\frac{{n - 2}}{{n - 3}} - \frac{1}{{n - 3}}} \right) + \frac{{n - 1}}{{n - 2}}\)
\( = n - 3 + \frac{{n - 1}}{{n - 2}} = n - 3 + \frac{{n - 2}}{{n - 2}} + \frac{1}{{n - 2}} = n - 2 + \frac{1}{{n - 2}}\)

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Let the three terms in A.P. be a – d, a, a + d.
given that a – d + a + a + d = 21 a = 7
then the three term in A.P. are 7 – d, 7, 7 + d
According to given condition 9 – d, 9, 21 + d are in G.P.
(9)² = (9 – d) (21 + d)
81 = 189 + 9d – 21d – d²
81 = 189 – 12d – d²
d² + 12d – 108 = 0
(d – 6) (d + 18) = 0
We get, d = 6, –18
Putting d = 6 in the term 7 – d, 7, 7 + d we get 1, 7, 13.

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Let us take the reciprocal of the numbers and the numbers will be in AP. Hence
\(a + (n - 1)d = \frac{1}{m}\), \(a + (m - 1)d = \frac{1}{n}\) and \(a + (m + n - 1)d = x\)
We have to find \(x\) by eliminating \(a\) and \(d\). Subtract second equation from first and third from second,
\((n - m)d = \frac{1}{m} - \frac{1}{n}\) \( \Rightarrow d = \frac{1}{{mn}}\)
\(nd = x - \frac{1}{n} \Rightarrow x = \frac{1}{m} + \frac{1}{n} = \frac{{mn}}{{m + n}}\)

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Let the first term and common ratio are \(a\) and \(r\), then
\(\frac{a}{{1 - r}} = 2(a + ar)\)
\( \Rightarrow 2(1 + r)(1 - r) = 1\)
\( \Rightarrow 1 - {r^2} = \frac{1}{2}\) or \(r = \pm \frac{1}{{\sqrt 2 }}\)

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Given that \(a,\,b,\,c\) are in GP, hence \({b^2} = ac\)
Also (log a – log2b), (log2b – log3c) and (log3c – loga) are in AP, hence
\(2\left[ {\log 2b - \log 3c} \right] = (\log a - \log 2b) + (\log 3c - \log a)\)
\( \Rightarrow 3\log 2b = 3\log 2c\)
Hence \(2b = 3c\)or \(4{b^2} = 9{c^2}\)
\(4(ac) = 9{c^2} \Rightarrow a = \frac{{9c}}{4}\)
Hence \(a:b:c = \frac{9}{4}:\frac{3}{2}:1 = 9:6:4\)
Since the biggest side is 9 and \({9^2} > {6^2} + {4^2}\), hence the triangle is an obtuse angled triangle.

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The ratio of third and second term is \(\frac{{3x + 3}}{{2x + 2}} = \frac{3}{2}\)
Hence \(\frac{{2x + 2}}{x} = \frac{3}{2} \Rightarrow x = - 4\)
The series will be – 4, – 6, – 9, – 13.5

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Let the common difference is \(d\), then
\(\frac{8}{2}\left( {2 \times 3 + 7d} \right) = 2 \times \frac{5}{2}\left( {6 + 4d} \right)\)
Hence \(d = \frac{3}{4}\)

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Including the 4 geometric means, there are 6 terms in the GP, where the first term is 2 and the last term is 64.
\( \Rightarrow 2({r^5}) = 64\) or \(r = 2\)
The four geometric means are 4, 8, 16, 32.

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The numbers \(\frac{1}{a},\,\frac{1}{{{H_1}}},........\frac{1}{{{H_n}}},\,\frac{1}{b}\) are in AP, hence
\(\frac{1}{{{H_1}}} - \frac{1}{a} = \frac{1}{b} - \frac{1}{{{H_n}}} = d\) and \(\frac{1}{b} = \frac{1}{a} + (n + 1)d\)
Now \(\frac{{{H_1} + a}}{{{H_1} - a}} + \frac{{{H_n} + b}}{{{H_n} - b}} = \frac{{\frac{1}{a} + \frac{1}{{{H_1}}}}}{{\frac{1}{a} - \frac{1}{{{H_1}}}}} + \frac{{\frac{1}{b} + \frac{1}{{{H_n}}}}}{{\frac{1}{b} - \frac{1}{{{H_n}}}}}\)
\( = \frac{{\frac{1}{a} + \frac{1}{a} + d}}{{ - d}} + \frac{{\frac{1}{b} + \frac{1}{b} - d}}{d}\)
\( = - \frac{2}{{ad}} - 1 + \frac{2}{{bd}} - 1 = \frac{2}{d}\left( {\frac{1}{b} - \frac{1}{a}} \right) - 2\)
\( = \frac{2}{d}\left[ {(n + 1)d} \right] - 2 = 2n\)

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Let us take the reciprocal of the numbers and the numbers will be in AP. Hence
\(a + (p - 1)d = \frac{1}{q}\), \(a + (q - 1)d = \frac{1}{p}\) and \(a + (pq - 1)d = x\)
We have to find \(x\) by eliminating \(a\) and \(d\). Subtract second equation from first and third from second,
\((p - q)d = \frac{1}{q} - \frac{1}{p}\) \( \Rightarrow d = \frac{1}{{pq}}\)
\((q - pq)d = \frac{1}{p} - x\)
Putting the value of \(d\), we get \(x = 1\)

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The given series is \(\frac{{\sqrt 5 }}{3},\,\frac{{\sqrt 5 }}{4},\frac{{\sqrt 5 }}{5},........\frac{{\sqrt 5 }}{{13}}\)
So we just need to find the position of 13 in the sequence 3, 4, 5, 6……
The correct answer is 11.
\(\sin d\left( {{\rm{cosec}}\,{a_1}{\rm{cosec}}\,{a_2} + {\rm{cosec}}\,{a_2}{\rm{cosec}}\,{a_3} + ..... + {\rm{cosec}}\,{a_{n - 1}}{\rm{cosec}}\,{a_n}} \right)\) is equal to:

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\(d = {a_2} - {a_1} = {a_3} - {a_2} = {a_4} - {a_3} = ...\)
First term of the given series is \(\sin d({\rm{cosec}}\,{a_1}{\rm{cosec}}\,{a_2})\)
\( = \frac{{\sin ({a_2} - {a_1})}}{{\sin {a_1}\sin {a_2}}} =  \frac{{sin{a_2}\cos {a_1} - \cos {a_2}\sin {a_1}}}{{\sin {a_1}\sin {a_2}}} = \cot {a_1} - \cot {a_2}\)
Similarly,
\(\sin d({\rm{cosec}}\,{a_2}{\rm{cosec}}\,{a_3}) = \frac{{\sin ({a_3} - {a_2})}}{{\sin {a_2}sin{a_3}}} = \cot {a_2} - \cot {a_3}\)
Writing all the terms in this manner and adding them we get the sum as
\((\cot {a_2} - \cot {a_1}) + (\cot {a_3} - \cot {a_2}) + ...(\cot {a_n} - \cot {a_{n - 1}})\)
\( = \cot {a_1} - \cot {a_N}\)

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