JEE Main & Advanced Mathematics Sequence & Series Question Bank

done Arithmetico - Geometric Series, Method of difference Total Questions - 17

• question_answer1)
  If \[|x|\,<1\], then the sum of the series \[1+2x+3{{x}^{2}}+4{{x}^{3}}+...........\infty \] will be
  A)
  \[\frac{1}{1-x}\] done clear
  B)
  \[\frac{1}{1+x}\] done clear
  C)
  \[\frac{1}{{{(1+x)}^{2}}}\] done clear
  D)
  \[\frac{1}{{{(1-x)}^{2}}}\] done clear
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• question_answer2)
  \[1+\frac{3}{2}+\frac{5}{{{2}^{2}}}+\frac{7}{{{2}^{3}}}+......\,\infty \,\]is equal to [DCE 1999]
  A)
  3 done clear
  B)
  6 done clear
  C)
  9 done clear
  D)
  12 done clear
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• question_answer3)
  The sum of infinite terms of the following series \[1+\frac{4}{5}+\frac{7}{{{5}^{2}}}+\frac{10}{{{5}^{3}}}+.........\] will be [MP PET 1981; RPET 1997; Roorkee 1992; DCE 1996, 2000]
  A)
  \[\frac{3}{16}\] done clear
  B)
  \[\frac{35}{8}\] done clear
  C)
  \[\frac{35}{4}\] done clear
  D)
  \[\frac{35}{16}\] done clear
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• question_answer4)
  The sum of the series \[1+3x+6{{x}^{2}}+10{{x}^{3}}+........\infty \]  will be
  A)
  \[\frac{1}{{{(1-x)}^{2}}}\] done clear
  B)
  \[\frac{1}{1-x}\] done clear
  C)
  \[\frac{1}{{{(1+x)}^{2}}}\] done clear
  D)
  \[\frac{1}{{{(1-x)}^{3}}}\] done clear
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• question_answer5)
  \[1+3+7+15+31+..........\]to \[n\]  terms = [IIT 1963]
  A)
  \[{{2}^{n+1}}-n\] done clear
  B)
  \[{{2}^{n+1}}-n-2\] done clear
  C)
  \[{{2}^{n}}-n-2\] done clear
  D)
  None of these done clear
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• question_answer6)
  \[2+4+7+11+16+......\]to \[n\]  terms = [Roorkee 1977]
  A)
  \[\frac{1}{6}({{n}^{2}}+3n+8)\] done clear
  B)
  \[\frac{n}{6}({{n}^{2}}+3n+8)\] done clear
  C)
  \[\frac{1}{6}({{n}^{2}}-3n+8)\] done clear
  D)
  \[\frac{n}{6}({{n}^{2}}-3n+8)\] done clear
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• question_answer7)
  Sum of n  terms of series \[12+16+24+40+.....\] will be [UPSEAT 1999]
  A)
  \[2\,({{2}^{n}}-1)+8n\] done clear
  B)
  \[2({{2}^{n}}-1)+6n\] done clear
  C)
  \[3({{2}^{n}}-1)+8n\] done clear
  D)
  \[4({{2}^{n}}-1)+8n\] done clear
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• question_answer8)
  If \[a,\ b,\ c\] are in H.P., then which one of the following is true [MNR 1985]
  A)
  \[\frac{1}{b-a}+\frac{1}{b-c}=\frac{1}{b}\] done clear
  B)
  \[\frac{ac}{a+c}=b\] done clear
  C)
  \[\frac{b+a}{b-a}+\frac{b+c}{b-c}=1\] done clear
  D)
  None of these done clear
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• question_answer9)
  The sum of the series \[1+\frac{1.3}{6}+\frac{1.3.5}{6.8}+....\infty \]is [UPSEAT 2001]
  A)
  1 done clear
  B)
  0 done clear
  C)
  \[\infty \] done clear
  D)
  4 done clear
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• question_answer10)
  \[{{n}^{th}}\] term of the series \[2+4+7+11+.......\]will be   [Roorkee 1977]
  A)
  \[\frac{{{n}^{2}}+n+1}{2}\] done clear
  B)
  \[{{n}^{2}}+n+2\] done clear
  C)
  \[\frac{{{n}^{2}}+n+2}{2}\] done clear
  D)
  \[\frac{{{n}^{2}}+2n+2}{2}\] done clear
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• question_answer11)
  The sum of the series \[1+2x+3{{x}^{2}}+4{{x}^{3}}+.........\]upto \[n\] terms is
  A)
  \[\frac{1-(n+1){{x}^{n}}+n{{x}^{n+1}}}{{{(1-x)}^{2}}}\] done clear
  B)
  \[\frac{1-{{x}^{n}}}{1-x}\] done clear
  C)
  \[{{x}^{n+1}}\] done clear
  D)
  None of these done clear
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• question_answer12)
  The sum of the first \[n\] terms of the series \[\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+.........\] is [IIT 1988; MP PET 1996; RPET 1996, 2000; Pb. CET 1994; DCE 1995, 96]
  A)
  \[{{2}^{n}}-n-1\] done clear
  B)
  \[1-{{2}^{-n}}\] done clear
  C)
  \[n+{{2}^{-n}}-1\] done clear
  D)
  \[{{2}^{n}}-1\] done clear
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• question_answer13)
  The sum of \[1+\frac{2}{5}+\frac{3}{{{5}^{2}}}+\frac{4}{{{5}^{3}}}+...........\]upto \[n\] terms is [MP PET 1982]
  A)
  \[\frac{25}{16}-\frac{4n+5}{16\times {{5}^{n-1}}}\] done clear
  B)
  \[\frac{3}{4}-\frac{2n+5}{16\times {{5}^{n+1}}}\] done clear
  C)
  \[\frac{3}{7}-\frac{3n+5}{16\times {{5}^{n-1}}}\] done clear
  D)
  \[\frac{1}{2}-\frac{5n+1}{3\times {{5}^{n+2}}}\] done clear
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• question_answer14)
  \[{{2}^{1/4}}{{.4}^{1/8}}{{.8}^{1/16}}{{.16}^{1/32}}..........\]is equal to  [MNR 1984; MP PET 1998; AIEEE 2002]
  A)
  1 done clear
  B)
  2 done clear
  C)
  \[\frac{3}{2}\] done clear
  D)
  \[\frac{5}{2}\] done clear
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• question_answer15)
  The sum of \[i-2-3i+4+.........\]upto 100 terms, where \[i=\sqrt{-1}\] is
  A)
  \[50(1-i)\] done clear
  B)
  \[25i\] done clear
  C)
  \[25(1+i)\] done clear
  D)
  \[100(1-i)\] done clear
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• question_answer16)
  \[{{99}^{th}}\] term of the series \[2+7+14+23+34+.....\] is [Pb. CET 2003]
  A)
  9998 done clear
  B)
  9999 done clear
  C)
  10000 done clear
  D)
  100000 done clear
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• question_answer17)
  If the set of natural numbers is partitioned into subsets \[{{S}_{1}}=\left\{ 1 \right\},\ {{S}_{2}}=\left\{ 2,\ 3 \right\},\ {{S}_{3}}=\left\{ 4,\ 5,\ 6 \right\}\] and so on. Then the sum of the terms in \[{{S}_{50}}\] is
  A)
  62525 done clear
  B)
  25625 done clear
  C)
  62500 done clear
  D)
  None of these done clear
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