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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
Insert three geometric means between `1/3` and 432. |
| Answer» Correct Answer - 2,12,72 | |
| 52. |
Find the GM between the numbers (i) 5 and 125 (ii) 1 and `9/16` (iii) 0.15 and 0.0015 (iv) -8 and -2 (v) -6.3 and -2.8 (v) `a^(3)b` and `ab^(3)` |
| Answer» Correct Answer - (i) 25 (ii) `3/4` (iii) 0.015 (iv) -4 (v) -0.42 (vi) `a^(2)b^(2)` | |
| 53. |
Write the value of `2.1bar(34)` in the form of a simple fraction. |
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Answer» Correct Answer - `2113/990` `x=2.1343434... rArr 10x=21.343434...` and `1000x=2134.3434 ...` `:. 990x=(2134-21)=2113 rArr x=2113/990`. Hence, `2.bar(134)=2113/990`. |
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| 54. |
Find two positive numbers a and b whose AM and GM are 34 and 16 respectively. |
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Answer» We have `(a+b)/2=34` and `sqrt(ab)=16` `rArr a+b=68` and `ab =256` `rArr (a-b)=sqrt((a+b)^(2)-4ab)=sqrt((68)^(2)-4xx256)=sqrt(3600)= pm 60` `rArr a+b=68` and `a-b= pm 60` `rArr (a+b=68, a-b=60) or (a+b=68, a-b=-60)` `rArr (a=64, b=4) or (a=4, b=64)` Hence, the required numbers are `(a=64, b=4) or (a=4, b=64)`. |
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| 55. |
If `x ,y ,z`are distinct positive numbers, then prove that `(x+y)(y+z)(z+x)>8x y zdot` |
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Answer» Using `AM gt GM`, we have `(x+y)/2 gt sqrt(xy), (y+z)/2 gt sqrt(yz)` and `(z+x)/2 gt sqrt(zx)` `rArr x+y gt 2sqrt(xy), y+z gt 2sqrt(yz) and z+x gt 2sqrt(zx)` `rArr (x+y) (y+z) (z+x) gt 2 sqrt(xy)xx2 sqrt(yz)xx 2sqrt(zx)` `rArr (x+y)(y+z)(z+x) gt 8xyz`. Hence, `(x+y)(y+z)(z+x) gt 8xyz`. |
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| 56. |
The sum of an infinite geometric series is 15 and the sum of thesquares of these terms is 45. Find the series. |
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Answer» Let a be the first term and r be the common ratio.Then, the series is `a+ar+ar^(2)+...oo`. Now, `(a+ar+ar^(2)+...oo)=15 rArr a/((1-r))=15` ...(i) and `(a^(2)+a^(2)r^(2)+a^(2)r^(4)+...oo) rArr a^(2)/((1-r^(2)))=45`. ...(ii) On squaring both sides of (i), we get `a^(2)/((1-r)^(2))=225` ...(iii) On dividing (iii) by (ii), we get `a^(2)/((1-r)^(2))xx((1-r^(2)))/a^(2)=225/45 rArr (1+r)/(1-r)=5` `rArr 1+r=5-5r` `rArr 6r=4 rArr r=2/3`. Putting `r=2/3` in (i), we get `a=15xx(1-2/3)=(15xx1/3)=5`. Thus, `a=5` and `r=2/3`. Hence, the required series is `(5+10/3+20/9+40/27+...oo)`. |
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| 57. |
The sum of an infinite geometric series is 6. If its first terms is 2, find its common ratio. |
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Answer» Correct Answer - `2/3` `a/(1-r)=6 rArr 1/(1-r)=6rArr 1-r=1/3 rArr r=(1-1/3)=2/3`. |
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| 58. |
If a, b, c and d are in G.P. show that `(a^2+b^2+c^2)(b^2+c^2+d^2)=(a b+b c+c d)^2`. |
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Answer» Let r be the common ratio of the GP a, b, c, d. Then, `b=ar, c=ar^(2)` and `d=ar^(3)`. `:. LHS=(a^(2)+b^(2)+c^(2))(b^(2)+c^(2)+d^(2))` `=(a^(2)+a^(2)r^(2)+a^(2)r^(4))(a^(2)r^(2)+a^(2)r^(4)+a^(2)r^(6))` `=a^(4)r^(2) (1+r^(2)+r^(4))^(2)`. And, `RHS =(ab+bc+cd)^(2)=(a^(2)r+a^(2)r^(3)+a^(2)r^(5))^(2)` `=a^(4)r^(2) (1+r^(2)+r^(4))^(2)`. Hence, `(a^(2)+b^(2)+c^(2)) (b^(2)+c^(2)+d^(2))=(ab+bc+cd)^(2)`. |
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| 59. |
If a, b, c, d are in GP, prove that `(b-c)^(2)+(c-a)^(2)+(d-b)^(2)=(a-d)^(2)`. |
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Answer» Let r be the common ratio of the GP a, b, c, d. Then. `b=ar, c=ar^(2)` and `d=ar^(3)`. `:. LHS =(b-c)^(2)+(c-a)^(2)+(d-b)^(2)` `=(ar-ar^(2))^(2)+(ar^(2)-a)^(2)+(ar^(3)-ar)^(2)` `={a(r-r^(2))}^(2)+{a(r^(2)-1)}^(2)+{a(r^(3)-r)}^(2)` `=a^(2)(r-r^(2))^(2)+a^(2)(r^(2)-1)^(2)+a^(2)(r^(3)-r)^(2)` `=a^(2).{(r-r^(2))^(2)+(r^(2)-1)^(2)+(r^(2)-r)^(2)}` `=a^(2).{(r^(2)+r^(4)-2r^(3))+(r^(4)+1-2r^(2))+(r^(6)+r^(2)-2r^(4))}` `=a^(2).(r^(6)-2r^(3)+1)=a^(2) (1-r^(3))^(2)` `=(a-ar^(3))^(2)=(a-d)^(2)=RHS`. Hence, `(b-c)^(2)+(c-a)^(2)+(d-b)^(2)=(a-d)^(2)`. |
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| 60. |
If the `pth , qth` and `rth`terms of aG.P. are `a , b , c`respectively, prove that: `a^(q-r).b^(r-p)c^(p-q)=1.` |
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Answer» Let A be the first term and R be the common ratio of the given GP. Then, `T_(p)=a rArr a=AR^((p-1))` `T_(q)=b rArr b=AR^((q-1))` `T_(r)=c rArr c= AR^((r-1))`. `:. a^((q-r)).b^((r-p)).c^((p-q))` `={AR^((p-1))}^((q-r)).{AR^((q-1))}^((r-p)).{AR^((r-1))}^((p-q))` `=A^((q-r)).R^((p-1)(q-r)).A^((r-p)).R^((q-1)(r-p)).A^((p-q)).R^((r-1)(p-q))` `=A^({(q-r)+(r-p)+(p-q)}).R^({(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)})` `=A^(0).R^(p(q-r)+q(r-p)+r(p-q)+(r-q)+(p-r)+(q-p))` `=(1xxR^(0))=(1xx1)=1`. Hence, `a^((q-r)).b^((r-p)).c^((p-q))=1`. |
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| 61. |
If `p, q, r` are in AP then prove that pth, qth and rth terms of any GP are in GP. |
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Answer» We have, `2q=p+r`. Let `T_(p), T_(q), T_(r)` be the given terms of a GP with terms =A common ratio = R. Then, `(T_(q))^(2)={AR^((q-1))}^(2)=A^(2)R^((2q-2))` and `(T_(p)xxT_(r))= {AR^((p-1))xxAR^((r-1))}=A^(2)R^((p+r-2)=A^(2) R^((2q-2))`. `:. (T_(p))^(2)=(T_(p)xxT_(r))` and hence `T_(p), T_(q), T_(r)` are in GP. |
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| 62. |
Find the sum of 8 terms of the GP 3, 6, 12, 24, ... |
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Answer» Here `a=3, r=2 gt 1` and `n=8`. Using the formula, `S_(n)=(a(r^(n)-1))/((r-1))`, we get `S_(8)=(3xx(2^(8)-1))/((2-1))=3xx(256-1)=3xx255=765`. |
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| 63. |
(iii) If `a, b c` are respectively the `pth, qth` and `rth` terms of the given `G.P.` then show that `(q-r) log a + (r-p) log b + (p-q)log c = 0`, where `a, b, c > 0. ` |
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Answer» Let A be the first term and R be the common ratio of the given GP. Then, `a=AR^((p-1)) rArr log a = log A+(p-1) log R` ...(i) `b=AR^((q-1)) rArr log b = log A+(q-1) log R` ...(ii) `c=AR^((r-1)) rArr log c= log A+(r-1) log R`. ...(iii) `:. (q-r) log a+(r-p) log b+(p-q) log c` `=(q-r) log [AR^((p-1))]+(r-p) log [AR^((q-1))]+(p-q) log [AR^((r-1))]` `=(log A) {(q-r)+(r-p)+(p-q)}+(log R){(p-1)(q-r)+(q-1) (r-p)+(r-1)(p-q)}` `=(log A)xx0+(log R)xx0=0`. |
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| 64. |
Show that the ratio of the sum of first n terms of aG.P. to the sum of terms from `(n+1)^(t h)`to `(2n)^(t h)`term is `1/(r^n)`. 9873740001 |
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Answer» Required ratio `=S_(n) : (S_(2n)-S_(n))`. Now, `S_(n)=(a(1-r^(n)))/((1-r)) and S_(2n)=(a(1-r^(2n)))/((1-r))=(a(1-r^(n))(1+r^(n)))/((1-r))` `:. (S_(2n)-S_(n))=(a(1-r^(n))(1+r^(n)))/((1-r))-(a(1-r^(n)))/((1-r))=(a(1-r^(n))(1+r^(n)-1))/((1-r))=(a(1-r^(n))r^(n))/((1-r))`. Hence, the required ratio is `1/r^(n)`. |
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| 65. |
If `x=Sigma_(n=0)^(oo) cos^(2n) theta, y= Sigma_(n=0)^(oo) sin^(2n) phi` and `z= Sigma_(n=0)^(oo) cos^(2n) theta sin^(2n) phi`, where `0 lt theta lt phi lt pi/2` then prove that `xz+yz-z=xy`. |
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Answer» We have `x=sum_(n=0)^(oo) cos^(2n)theta=1+cos^(2) theta+ cos^(4) theta+...oo` `= a/((1-r))`, where `a=1` and `r= cos^(2) theta` [sum of an infinite GP] `=1/((1- cos^(2) theta))=1/(sin^(2) theta)`. `:. sin^(2) theta=1/x` ...(i) `y= sum_(n=0)^(oo) sin^(2n) phi=1 + sin^(2) phi + sin^(4) phi +...oo` `=a/((1-r))`, where `a=1` and `r= sin^(2) phi` [sum of an infinite GP] `= 1/((1-sin^(2) phi))=1/(cos^(2) phi)`. `:. cos^(2) phi =1/y` ...(ii) `z= sum_(n=0)^(oo) cos^(2n) theta sin^(2n) phi =1+ cos^(2) theta sin^(2) phi +cos^(4) theta sin^(4) phi +...oo` `=a/((1-r))`, where `a=1` and `r=cos^(2) theta sin^(2) phi` `1/((1-cos^(2) theta sin^(2) phi))=1/(1-(1-sin^(2) theta)(1- cos^(2) phi))` `=1/(1-(1-1/x)(1-1/y))=1/(1-(1-1/x-1/y+1/(xy)))` [using (i) and (ii)] `=1/((1/x+1/y+1/(xy)))=(xy)/((x+y-1))` `:. z=(xy)/(x+y-1) implies xz+yz-z=xy`. Hence, `xz+yz-z=xy`. |
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| 66. |
Find the sum of the grometric series `1+1/2+1/4+1/8+...` to 12 terms. |
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Answer» Here `a=1, r=1/2 lt 1` and `n=12`. Using the formula, `S_(n)=(a(1-r^(n)))/((1-r))`, we get `S_(12)=(1xx{1-(1/2)^(n)})/((1-1/2))=((1-1/2^(12)))/((1/2))=((2^(12)-1))/2^(11)=4095/2048`. |
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| 67. |
If a, b, c are in GP, prove that `a^(2), b^(2), c^(2)` are in GP. |
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Answer» Let `b=ar` and `c=ar^(2)`. Then `a^(2)c^(2)=a^(2)(ar^(2))^(2)=a^(2)xxa^(2)r^(4)=a^(4)r^(4)=(b^(2))^(2)`. |
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| 68. |
How many terms of the geometric series `1+4+16+64+`will make the sum `5461`? |
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Answer» Let the required number of terms be n. Here, `a=1, r=4 gt 1` and `S_(n)=5461`. `:. S_(n)=(a(r^(n)-1))/((r-1)) rArr (1xx(4^(n)-1))/((4-1))=5461` `rArr (4^(n)-1)=16383` `rArr 4^(n)=16384=4^(7)` `rArr n=7`. Hence, the required number of terms is 7. |
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| 69. |
Insert two number between 3 and 81 so that theresulting sequence is G.P. |
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Answer» Let the required numbers be `G_(1)` and `G_(2)`. Then, `3, G_(1), G_(2), 81` are in GP. Let the common ratio of this GP be r. Then, `T_(4)=81 rArr ar^((4-1))=81` `rArr 3xxr^(3)=81 rArr r^(3)=27=(3)^(3) rArr r=3`. `:. G_(1)=(3 xx r)(3xx3)=9` and `G_(2)=(3xxr^(2))=(3xx9)=27`. Hence, the required numbers are 3 and 27. |
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| 70. |
`3/5+4/5^2+ 3/5^3+4/5^4+ ... ` to 2n terms; |
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Answer» Correct Answer - `19/24.((5^(2n)-1)/5^(2n))` Given sum `={3/5+3/5^(3)+3/5^(5)+..."to n terms"}+{4/5^(2)+4/5^(4)+..."to n terms"}` `=(3/5xx{1-(1/5^(2))^(n)})/((1-1/5^(2)))+(4/5^(2)xx{1-(1/5^(2))^(n)})/((1-1/5^(5)))` `=(3/5xx25/24) ((5^(2n)-1))/5^(2n)+(4/25xx25/24) ((5^(2n)-1))/5^(2n)` `=(5/8+6/1)((5^(2n)-1)/5^(2n))=19/24.((5^(2n)-1)/5^(2n))`. |
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| 71. |
Evaluate : (i) `sum_(n=1)^(10)(2+3^(n))" "`(ii) `sum_(k=1)^(n)[2^(k)+3^((k-1))]" "`(iii) `sum_(n=1)^(8) 5^(n)` |
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Answer» Correct Answer - (i) 29544 (ii) `1/2[2^((n+2))+3^(n)-5]` (iii) 488280 (i) `sum_(n=1)^(10) (2+3^(n))=(2+3^(1))+(2+3^(3))+(2+3^(3))+...+(2+3^(10))` `=(2+2+2+..."taken 10 times")+(3+3^(2)+3^(3)+...+3^(10))` `=(2xx10)+{3xx((3^(10)-1)/(3-1))}` `=20+((59049-1)/2)=(20+29524)=29544`. (ii) `sum_(k=1)^(n)[2^(k)+3^((k-1))]=(2^(1)+3^(0))+(2^(2)+3^(1))+(2^(3)+3^(2))+`... up to n terms `=(2^(1)+2^(2)+2^(3)+..."to n terms")+(1+3+3^(2)+..."to n terms")` `=2((2^(n)-1)/(2-1))+1. ((3^(n)-1)/(3-1))` `=2(2^(n)-1)+1/2(3^(n)-1)=2^((n+1))-2+1/2 . 3^(n)-1/2` `=2^((n+1))+1/2. 3^(n)-5/2=2^((n+1))+1/2 (3^(n)-5)=1/2 [2^((n+2))+3^(n)-5]`. (iii) `sum_(n=1)^(8) 5^(n)=[5+5^(2)+5^(3)+..."to 8 terms"]` `=5((5^(8)-1)/(5-1))=(5(390625-1))/4=(5xx390624)/4` `=(5xx97656)=488280`. |
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| 72. |
Find the sum of the series : (i) `8+88+888+` ... to n terms (ii) `3+33+333+` ... to n terms (iii) `0.7+0.77+0.777+`... to n terms |
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Answer» Correct Answer - (i) `8/81[10^((n+1))-10-9n]` (ii) `1/27 [10^((n+1))-10-9n]` (iii) `7/81(9n-1+1/10^(n))` (i) Given sum `=8xx[1+11+111+..."to n terms"]` `=8/9xx[9+99+999+..."to n terms"]` `=8/9xx[(10-1)+(10^(2)-1)+(10^(3)-1)+..."to n terms"]` `=8/9xx{(10+10^(2)+10^(3)+..."to n terms")-n}` `=8/9 xx{10xx((10^(n)-1)/(10-1))-n}=8/81 [10^((n-1))-10-9n]`. (iii) Given series `=7/10+77/100+777/1000+..."to n terms"` `=7xx{1/10+11/100+111/1000+..."to n terms"}` `=7/9xx{9/10+99/100+999/1000+..."to n terms"}` `=7/9xx{(1-1/10)+(1-1/10^(2))+(1-1/10^(3))+..."to n terms"}` `7/9xx{n-(1/10+1/10^(2)+1/10^(3)+..."to n terms")}` `=7/9xx{n-(1/10(1-1/10^(n)))/((1-1/10))}=7/81xx(9n-1+1/10^(n))`. |
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| 73. |
A. G.P. consists of an even number of terms. If the sum of all theterms is 5 times the sum of the terms occupying he odd places. Find thecommon ratio of the G.P. |
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Answer» Correct Answer - `r=4` Let us consider a GP with first term a and common ratio r and having 2n terms. Then, `T_(1)+T_(2)+T_(3)+T_(4)+...+T_(2n)=5(T_(1)+T_(3)+T_(5)+...+T_(2n-1))` `rArr a+ar+ar^(2)+...+ar^((2n-1))=5[a+ar^(2)+ar^(4)+...+ar^((2n-2))]` `rArr a((1-r^(2n))/(1-r))=5a [(1-(r^(2))^(n))/(1-r^(2))] rArr 1+r=5 rArr r=4`. |
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| 74. |
Show that the sequence given by `T_(n)=(2xx3^(n))` for all `n in N` is a GP. Find its first term and the common ratio. |
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Answer» We have `T_(n)=(2xx3^(n))` and `T_(n+1)=(2xx3^(n+1))`. `:. T_(n+1)/T_(n)=(2xx3^(n+1))/(2xx3^(n))=3` (constant) for all `n in N`. Also, `T_(1)=(2xx3^(1))=(2xx3)=6` `:.` the first term `=6` and the common ratio `=3`. |
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| 75. |
The sum of an infinite GP is `80/9` and its common ratio is `(-4)/5`. Find its first term. |
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Answer» Let a be the first term of the given infinite GP. Here, `r=(-4)/5` and `|r|=|(-4)/5|=4/5 lt 1`. The sum of this infinite GP is `S=80/9`. `:. S=a/((1-r)) rArr a/((1+4/5))=80/9` `rArr (5a)/9=80/9 rArr 5a=80 rArr a=16`. Hence, the first term is 16. |
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| 76. |
The sum of n terms of a progression is `(2^(n)-1)`. Show that it is a GP and find its common ratio. |
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Answer» Correct Answer - `r=2` `S_(n)=(2^(n)-1)` and `S_(n-1)=[2^((n-1))-1]`. `:. T_(n)=(S_(n)-S_(n-1))=[2^(n)-2^((n-1))]rArr T_(1)=1, T_(2)=2, T_(3)=4, T_(4)=8, ...` Clearly, 1, 2, 4, 8, ... is a GP with common ratio =2 |
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| 77. |
If the 4th, 10th and 16th terms of a GP are x, y, z respectively, prove that x, y, z are in GP. |
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Answer» Let a be the first term and r be the common ratio of the given GP. Then, `T_(4)=x, T_(10)=y` and `T_(16)=z`. `rArr ar^(3)=x, ar^(9)=y` and `ar^(15)=z`. `:. y^(2)=(ar^(9))^(2)=a^(2)r^(18)` and `xz=(ar^(3))(ar^(15))=a^(2)r^(18)`. Consequently, we have `y^(2)=xz`. Hence, x, y, z are in GP. |
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| 78. |
The 2nd and 5th terms of a GP are `(-1)/2` and `1/16` respectively. Find the sum of the GP up to 8 terms. |
| Answer» Correct Answer - `85/128` | |
| 79. |
The first term of a GP is 27 and its 8th terms is `1/81`. Find the sum of its first 10 terms. |
| Answer» Correct Answer - `81/2 (1-1/3^(10))` | |
| 80. |
In a GP `a_1=3,a_n=96` and `S_n=189` .Find the number of terms |
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Answer» Here, `a=3, l=96` and `S_(n)=189`. Let the common ratio of the given GP be r. Then, `S_(n)=((lr-a))/((r-1)) rArr ((96r-3))/((r-1))=189` `rArr (96r-3)=(189r-189)` `rArr 93r=186 rArr r=2`. Now, `l=ar^(n-1) rArr 3xx2^(n-1)=96` `rArr 2^(n-1)=32=2^(5) rArr n-1=5 rArr n=6`. Hence, `n=6`. |
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| 81. |
Three years before the population of a village was 10000. If at the end of each year, 20% of the prople migrated to a neaby town, what is its present population ? |
| Answer» Correct Answer - 5120 | |
| 82. |
What will Rs. 5000 amount to in 10 years, compounded annually at 10 % per annume ? `["Given "(1.1)^(10)=2.594]` |
| Answer» Correct Answer - Rs. 12970 | |
| 83. |
What will Rs. 10000 amount to in 4 years to in 4 years after its deposit in a bank which pays annual interest at the rate of 10% per annum compounded annually ? |
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Answer» Original sum = Rs. 10000. Amount after 1 year = Rs. `10000+Rs. (10000xx10/100xx1)= Rs. 11000`, amount after 2 years `=Rs. 11000 + Rs. 11000+Rs. (11000xx10/100xx1)= Rs. 12100`, amount after 3 years `= Rs. 12100+ Rs. (12100xx10/100xx1)= Rs. 13310`, and so on. Thus, these amounts from a GP `10000, 11000, 12100, 13310, ...` such that `11000/10000=12100/11000=13310/12100=11/10`. In this GP, we have `a=10000, r=11/10`. Amount after 4 years `=T_(5)=ar^((5-1))=ar^(4)` `=Rs. [10000xx(11/10)^(4)]= Rs. 14641`. Hence, the amount after 4 years is Rs. 14641. Alternative method Here, `P=Rs. 10000, R=10% p.a. and T=4` years. `:.` amount after 4 years `=Rs. {Pxx(1+R/100)^(T)}` `= Rs. {10000xx(1+10/100)^(4)}` `=Rs. {10000xx(11/10)^(4)}= Rs. 14641`. Hence, the amount after 4 years is Rs. 14641. |
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| 84. |
If `(a^(2)+b^(2)), (ab+bc), (b^(2)+c^(2))` are in GP then prove that a, b, c are also in GP. |
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Answer» Let `(a^(2)+b^(2)), (ab+bc)` and `(b^(2)+c^(2))` be in GP. Then, `(ab+bc)^(2)=(a^(2)+b^(2))(b^(2)+c^(2))` `rArr a^(2)b^(2)+b^(2)c^(2)+2ab^(2)c=a^(2)b^(2)+a^(2)c^(2)+b^(4)+b^(2)c^(2)` `rArr b^(4)+a^(2)c^(2)-2ab^(2)C=0` `rArr (b^(2)-ac)^(2)=0 rArr b^(2)-ac=0 rArr b^(2)=ac`. Hence, a, b, c are in GP. |
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| 85. |
(ii)If the third term of G.P.is `4`, then find the product of first five terms |
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Answer» Correct Answer - 1024 Let a be the first term and r be the common ratio of the GP. Then, the first five terms are `a, ar, ar^(2), ar^(3), ar^(4)`. Also, `T_(3)=4 rArr ar^(2) =4`. Product of first five terms `=(axxarxxar^(2)xxar^(3)xxar^(4))=a^(5)r^(10)=(ar^(2))^(5)=(4)^(5)=1024`. |
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| 86. |
The three numbers are in A.P and their sum is 21. If the first and second are decrease by 1 each and third is increased by 7, they form a G.P Find the numbers of A.P. |
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Answer» Correct Answer - `(12,7,2) or (3,7,11)` Let the required numbers be `(a-d), a, (a+d)`. Then, `3a=21 rArr a=7`. So, these numbers are `(7-d), 7, (7+d)`. `:. (7-d), 6` and `(8+d)` are in GP. So, `6/((7-d))=((8+d))/6 rArr (7-d) (8+d)=36 rArr d^(2)+d-20=0` `rArr (d+5) (d-4)=0 rArr d= -5 or d=4`. Hence, the required numbers are `(12, 7, 2) or (3, 7, 11)`. |
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| 87. |
Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the `4^(t h)`by 18. |
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Answer» Let the required numbers be `a, ar, ar^(2)` and `ar^(3)`. Then, `T_(3)-T_(1)=9rArr ar^(2)-a=9rArr a(r^(2)-1)=9` ...(i) and `T_(2)-T_(4)=18 rArr ar-ar^(3)=18 rArr ar(1-r^(2))=18` ...(ii) On dividing (ii) by (i), we get `r=-2`. Putting `r=-2` in (i), we get `a(4-1)=9 rArr a=3`. Hence, the required numbers are 3, -6, 12 and -24. |
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| 88. |
The product of three numbers in GP is 1000. If 6 is added to the second number and 7 is added to the third number, we get an AP. Find the numbers. |
| Answer» Correct Answer - `5, 10, 20 or 20, 10, 5` | |
| 89. |
The number of bacteria in a certain culture doubles every hour. Ifthere were 30 bacteria present in the culture originally, how many bacteriawill be present at the end of 2nd hour, 4th hour andnth hour? |
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Answer» The bacteria present in the culture originally, at the end of 1st hour, at the end of 2nd, hour, at the end of 3rd hour and so on, are 30, 60, 120, 240,... this is a GP with `a=30` and `r=60/30=2`. Number of bacteria at the end of 2nd hour `=T_(3)=ar^(2)=(30xx2)^(2)=120`. Number of bacteria at the end of 4th hour `=T_(5)=ar^(4)=(30xx2^(4))=480`. Number of bacteria at the end of nth hour `=T_(n+1)=ar^((n+1-1))=30xx2^(n)`. |
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| 90. |
The sum of three numbers in GP is `39/10` and their product is 1. Find the numbers |
| Answer» Correct Answer - `5/2, 1, 2/5 or 2/5, 1, 5/2` | |
| 91. |
If `a ,b ,c ,d`are in G.P., prove that `a+b+,b+c ,c+d`are also in G.P. |
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Answer» Let a, b, c, d be in GP with common ratio r. Then, `b=ar, c=ar^(2)` and `d=ar^(3)`. `:. (a+b)=(a+ar)=a(1+r), (b+c)=(ar+ar^(2))=ar(1+r)`, `(c+d)=(ar^(2)+ar^(3))=ar^(2) (1+r)`. `:. (b+c)^(2)=a^(2)r^(2)(1+r)^(2)` and `(a+b)(c+d)=a^(2)r^(2)(1+r)^(2)`. Consequently, `(b+c)^(2)=(a+b)(c+d)`. Hence, `(a+b), (b+c)` and `(c+d)` are in GP. |
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| 92. |
The inventor of the chess board suggested a reward of one grain of whenfor the first square, 2 grains for the second, 4 grains for the third and soon, doubling the number of the grains for subsequent squares. How many grainswould have to be given to inventory? (There are 64 square sin the chessboard). |
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Answer» Required numbers of grains `=1+2^(1)+2^(2)+2^(3)+...` to 64 terms `=1+(2^(1)+2^(2)+2^(3)+...2^(63))` `=1+(2(2^(63)-1))/((2-1))=(1+2^(64)-2)=(2^(64)-1)`. |
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| 93. |
If `a , b , c`are in G.P. prove that `loga , logb , logc`are in A.P. |
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Answer» Let a, b, c be in GP. Then, `b^(2)=ac rArr log b^(2) = log (ac)` `rArr 2 log b = log a+log c` `rArr log a, log b, log c` are in AP. |
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| 94. |
Three numbers whose sum is 15 are in A.P. If they are added by 1,4 and 19 respectively, they are in GP. Thenumbers are |
| Answer» Correct Answer - `(2, 5, 8) or (26, 5, -16)` | |
| 95. |
The sum of three numbers in GP. Is 56. If we subtract 1, 7, 21 fromthese numbers in that order, we obtain an arithmetic progression. Find thenumbers. |
| Answer» Correct Answer - `8,16,32` | |
| 96. |
the lengths of three unequal egdes of a rectangular solids block are in GP .if the volume of the block is 26 `cm^(3)` and the total surface area is `252cm ^(2)` then the length of the longest edge is |
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Answer» Let the lengths of its edges be `a/r` cm, a cm and ar cm. Then, its volume `=(a/rxxaxxar) cm^(3)=a^(3) cm^(3)`. `:. a^(3)=216=6^(3) rArr a=6`. So, the edges are 6r cm, 6 cm and `6/r` cm. `:.` surface area `=2[lb+bh+lh]` `=2[6rxx6+6xx6/r+6rxx6/r] cm^(2)` `=2[36r+36/r+36] cm^(2) =72 xx[r+1/r+1] cm^(2)`. `:. 72 xx(r+1/r+1)=252` `rArr 2(r^(2)+1+r)=7r rArr 2r^(2)-5r+2 =0` `rArr (r-2) (2r-1) =0 rArr r=2` or `r=1/2`. Each value of r gives the longest edge `=12` cm. |
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| 97. |
Find the 7th and nth terms of the GP `0.4, 0.8, 1.6, ...` . |
| Answer» Correct Answer - `25.6, 2^(n)/5` | |