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InterviewSolution
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.
| 1. |
If `(a+b x)/(a-b x)=(b+c x)/(b-c x)=(c+dx)/(c-dx)(x!=0)`, then show that `a , b , c a n d d`are in G.P. |
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Answer» `(a+bx)/(a-bx)=(b+cx)/(b-cx) rArr ((a+bx)+(a-bx))/((a+bx)-(a-bx))=((b+cx)+(b-cx))/((b+cx)-(b-cx))rArr b/a=c/b`. Similarly, `(b+cx)/(b-cx)=(c+dx)/(c-dx)rArr c/b=d/c`. `:. b/a=c/b=d/c rArr a, b, c, d` and in GP. |
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| 2. |
If p, q, r are in G.P. and the equations, `p x^2+2q x+r=0`and `dx^2+2e x+f=0`have a common root, then show that `d/p`,`e/q`,`f/r`are in A.P. |
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Answer» Since, p, q, r are in GP, we have `q^(2)=pr` ...(i) On solving `px^(2)+2qx+r=0`, we get `x=(-2q pm sqrt(4q^(2)-4pr))/(2p)=(-2q)/(2p)=(-q)/p` [using (i)] Thus, `x=(-q)/p` is a repeated root of `px^(2)+2pr+r=0`. `:. x=(-q)/p` is also a root of `dx^(2)+2ex+f=0` `rArr d.((-q)/p)^(2)+2e((-q)/p)+f=0` `rArr dq^(2)-2eqp+fp^(2)=0` ...(ii) `rArr d/p-(2e)/q+(fp)/q^(2)=0` [on dividing (ii) by `pq^(2)`] `rArr d/p+f/r=(2e)/q`. Hence, `d/p, e/q, f/r` are in AP. |
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| 3. |
If `a a n d b`are the roots of `x^2-3x+p=0 a n d c , d`are the roots `x^2-12 x+q=0`where `a , b , c , d`form a G.P. Prove that `(q+p):(q-p)=17 : 15.` |
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Answer» Let r be the common ratio of the given GP. Then, `b=ar, c=ar^(2)` and `d=ar^(3)`. Also, `a+b=3, ab=p, c+d=12` and `cd=q`. Now, `a+b=3, c+d=12 rArr a(1+r)=3` and `ar^(2) (1+r)=12` `rArr (ar^(2) (1+r))/(a(1+r))=4rArr r^(2) = 4 rArr r=2` `rArr a(1+2)=3rArr a=1`. `:. p=ab=a xx ar=a^(2) r=(1^(2) xx2)=2` and `q=cd=ar^(2) xxar^(3)=a^(2) r^(5)=(1)^(2)xx2^(5)=32`. Hence, `(q+p)/(q-p)=(32+2)/(32-2)=34/30=17/15`. |
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| 4. |
If `a , b , c`are in A.P. `b , c , d`are in G.P. and `1/c ,1/d ,1/e`are in A.P. prove that `a , c , e`are in G.P.? |
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Answer» a, b, c are in `AP rArr 2b = a+c` ...(i) b, c, d are in GP `rArr c^(2)=bd` ...(ii) `1/c, 1/d, 1/e` are in `APrArr 2/d =(1/c+1/e)=((c+e))/(ce)` `rArr d=(2ce)/((c+e))` ...(iii) Now, `c^(2)=bd rArr c^(2) =((a+c))/2. (2ce)/((c+e))` [from (i) and (iii)] `rArr c=((a+c)e)/((c+e))` `rArr c(c+e)=(a+c)e` `rArr c^(2)=ae` `rArr` a, c, e are in GP. Hence, a, c, e are in GP. |
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| 5. |
If `a ,b ,c`are in G.P. then prove that `(a^2+a b+b^2)/(b c+c a+a b)=(b+a)/(c+b)` |
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Answer» Let r be the common ratio of the given GP. Then, `b=ar and c=ar^(2)`. `:. LHS =(a^(2)+a^(2)r+a^(2)r^(2))/(a^(2)r+a^(2)r^(3)+a^(2)r^(2))=(a^(2)(1+r+r^(2)))/(a^(2)r(1+r+r^(2)))=1/r` `RHS=(ar+a)/(ar^(2)+ar)=(a(1+r))/(ar(1+r))=1/r`. Hence, `LHS=RHS`. |
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| 6. |
If `S`be the sum, `P`the product and `R`the sum of the reciprocals of `n`terms of a G.P. prove that `(S/R)^n=P^2dot` |
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Answer» Let the given GP be `a, ar, ar^(2),..., ar^((n-1))`. Then, `S=(a(1-r^(n)))/((1-r))` ..(i) and `P=[axxarxxar^(2) xx...xxar^((n-1))]` `rArr P=a^(n)r^([1+2+3+...+(n-1)])=a^(n) r^(1/2 (n-1) n)` `rArr P=a^(n)r^(1/2 (n-1) n)` `rArr P^(2)=a^(2n)r^((n-1)n)` ...(ii) And, `R={1/a+1/(ar)+...+1/(ar^(n-1))}=(1/a). ({1/r^(n)-1})/({1/r-1})` `rArr R=(r/a). ((1-r^(n)))/((1-r).r^(n))` `rArr R=((1-r^(n)))/(a(1-r).r^((n-1)))`. ...(iii) On dividing (i) by (iii), we get `S/R=(a(1-r^(n)))/((1-r)). (a(1-r).r^((n-1)))/((1-r^(n)))=a^(2) r^((n-1))` `:. (S/R)^(n)={a^(2)r^((n-1))}^(n)=a^(2n).r^((n-1)n)=P^(2)` [using (ii)]. Hence, `P^(2)=(S/R)^(n)`. |
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| 7. |
If `(a-b), (b-c),(c-a)`are in G.P. then prove that `(a+b+c)^2=3(a b+b c+c a)` |
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Answer» `((b-c))/((a-b))=((c-a))/((b-c)) rArr (b-c)^(2)=(a-b)(c-a)` `rArr b^(2)+c^(2)-2bc=ac-a^(2)-bc+ab` `rArra^(2)+b^(2)+c^(2)=ab+bc+ac` `rArr (a+b+c)^(2)=3(ab+bc+ac)" "["adding "2(ab+bc+ac)" on both sides"]`. |
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| 8. |
Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that `P^2R^n=S^n`. |
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Answer» Let the three terms of the given GP be `a/r`, a and ar. Then, `S=(a/r+a+ar)=a((1+r+r^(2))/r), P=(a/rxxaxxar)=a^(3)` and `R=(r/a+1/a+1/(ar))=1/a(r+1+1/r)=1/a((r^(2)+r+1)/r)=1/a((1+r+r^(2))/r)`. `:. P^(2)R^(3)=a^(6) xx((1+r+r^(2))^(3))/(a^(3)r^(3))=a^(3)((1+r+r^(2))/r)^(3)=S^(3)` Hence, `P^(2)R^(3)=S^(3)`. |
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| 9. |
Which of the foliowing statement(s) is/are true? (A) If `a^x =b^y=c^z and a,b,c` are in GP, then x, y, z are in HP (B) If `a^(1/x)=b^(1/y)=c^(1/z)` and `a,b,c` are in GP, then `x,y,z` are in AP |
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Answer» Let `a^(1//x)=b^(1//y)=c^(1//z)=m` Then, `a=m^(x), b=m^(y)` and `c=m^(z)`. Since a, b, c are in GP we have `b^(2)=ac rArr (m^(y))^(2)=(m^(x)xxm^(z))` `rArr m^(2y)=m^(x+z) rArr 2y=x+z`. `:. x, y, x` are in AP. |
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| 10. |
Find the sum of each of the following infinite series : `6+1.2+0.24+oo` |
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Answer» Correct Answer - 7.5 `a=6 and r=1.2/6=0.2` `:. S=a/((1-r))=6/0.8=60/8=15/2=7.5` |
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| 11. |
If a, b, c are in GP, prove that `1/((a+b)), 1/(2b), 1/(b+c)` are in AP. |
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Answer» Let `b=ar` and `c=ar^(2)`. Then, `1/(a+b)=1/(a+ar)=1/(a(1+r)), 1/(2b)=1/(2ar), and 1/((b+c))=1/(ar(1+r))` `:. 1/((a+b))+1/((b+c))=1/(a(1+r))=1/(ar(1+r))=((1+r))/(ar(1+r))=1/(ar)=2xx(1/(2b))`. Hence, `1/((a+b)), 1/(2b), 1/((b+c))` are in AP. |
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| 12. |
If `a, b, c` are in AP or GP or HP, then `(a-b)/(b-c)` is equal to |
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Answer» For a, b, c to be in GP, we must have `b/a=c/b=r` or `b/a=c/b=1/r`. Case I When `b/a=c/b=r`. In this case, `b=ar` and `c=br`. `:. (a-b)/(b-c)=(a-ar)/(b-br)=(a(1-r))/(b(1-r))=a/b` Case II When `b/a=c/b=1/r`. In this case, `a=br` and `b=cr` `:. (a-b)/(b-c)=(br-b)/(cr-c)=(b(r-1))/(c(r-1))=b/c`. Hence, the value of `((a-b)/(b-c))` is `a/b` or `b/c`. |
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| 13. |
Find the sum of each of the following infinite series : `sqrt(2)-1/sqrt(2)+1/(2sqrt(2))-1/(4sqrt(2))+...oo` |
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Answer» Correct Answer - `8(2+sqrt(2))` `a=8 and r=(4 sqrt(2))/8=1/sqrt(2)` `:. S=a/((1-r))=8/((1-1/sqrt(2)))=(8sqrt(2))/((sqrt(2)-1))xx((sqrt(2)+1))/((sqrt(2)+1))=8(2+sqrt(2))`. |
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| 14. |
If `A=1+r^a+r^(2a)+`to `ooa n dB=1+r^b+r^(2b)+oo`, prove that`r=((A-1)/A)^(1//a)=((B-1)/B)^(1//a)` |
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Answer» By summing the given infinite geometric series, we get `x=1/((1-r^(a))) and y=1/((1-r^(b)))` `rArr (1-r^(a))=1/x` and `(1-r^(b))=1/y` `rArr r^(a)=(1-1/x) and r^(b)=(1-1/y)` `rArr r=((x-1)/x)^(1/a) and r=((y-1)/y)^(1/b)`. Hence, `r=((x-1)/x)^(1/a)=((y-1)/y)^(1/b)`. |
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| 15. |
9. Show that the product of n geometric means between a and b is equal to the nth power of the single GM between a and b. |
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Answer» Let `G_(1), G_(2), ... G_(n)` be n GMs between a and b, and let r be its common ratio. Then, `a, G_(1), G_(2), ..., G_(n), b` are in GP. `:. T_(n+2)=b rArr ar^((n+2-1))=b rArr r^((n+1))=(b/a) rArr r=(b/a)^(1/((n+1)))` ...(i) `:. G_(1)xxG_(2)xxG_(3)xx...xxG_(n)=(ar)xx(ar^(2))xx...xx(ar^(n))` `=a^(n)xxr^((1+2+...+n))=a^(n)xxr^(1/2n (n+1))` `=a^(n)xx(b/a)^(n//2)=a^(n//2)xxb^(n//2)` [using (i)] `=(sqrt(ab))^(n)=G^(n)`, where `G=sqrt(ab)`. |
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| 16. |
Find the sum of the following series:`1/2+1/(3^2)+1/(2^3)+1/(3^4)+1/(2^5)+1/(3^6)+oo` |
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Answer» The given series can be expressed as the sum of teo infinite geometric series, shown below. `{1/2 +1/3^(2)+1/2^(3)+1/3^(4)+1/2^(5)+1/3^(6)+...oo}` `={1/2+1/2^(3)+1/2^(5)+...oo}+{1/3^(2)+1/3^(4)+1/3^(6)+..oo}` `{"an infinite GP with "a=1/2 and r=1/4}` `{"an infinite GP with "a=1/9 and r=1/9}` `=((1//2))/((1-1/4))+((1//9))/((1-1/9))=((1//2))/((3//4))+((1//9))/((8//9))` `=(1/2xx4/3)+(1/9xx9/8)=(2/3+1/8)` `=(16+3)/24=19/24`. |
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| 17. |
If a, b, c are in AP and a, b, d are in GP, show that `a, (a-b)` and `(d-c)` are in GP. |
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Answer» `(a+c)=2b and b^(2)=ad`. We have to prove that `(a-b)^(2)=a(d-c)`. Now, `(a-b)^(2)=a^(2)+b^(2)-2ab=a^(2)+ad-a(a+c)` `rArr (a-b)^(2)=ad-ac=a(d-c)`. Hence, `a(a-b), (d-c)` are in GP. |
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| 18. |
If a, b, c are in AP, and a, x, b and b, y, c are in GP then show that `x^(2), b^(2), y^(2)` are in AP. |
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Answer» `x^(2)=ab, y^(2)=bc and a+c=2b`. `:. x^(2)+y^(2)=(ab+bc)=b(a+c)=2b^(2)`. Hence, `x^(2), b^(2), y^(2)` are in AP. |
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| 19. |
Find the rational number whose decimal form is `1.bar(345)`. |
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Answer» We have `1.bar(345)=1.3454545...oo` `=1.3+0.045+0.00045+...oo` `=1.3+{45/10^(3)+45/10^(5)+...oo}` `=1.3+((45/10^(3)))/((1-1/10^(2)))" "[("this being a GS in which"),(a=45/10^(3) and r=1/10^(2) lt 1)]` `=1.3+45/1000xx100/99=13/10+45/990=(13/10+1/22)` `=(143+5)/110=148/110=74/55`. Hence, `1.bar(345)=74/55`. |
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| 20. |
Find the rational number whose decimal form is `0.bar(142)`. |
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Answer» We have `0.bar(142).0.142424242...` `=0.1+0.042+0.00042+0.0000042+...oo` `=1/10+{42/10^(3)+42/10^(5)+42/10^(7)+...oo}` `=1/10+((42/10^(3)))/((1-1/10^(2)))[("this being a GS in which "),(a=42/10^(3) and r=(42/10^(5)xx10^(3)/42)=1/10^(2) lt 1)]` `=1/10+(42/1000xx100/99)=(1/10+42/990)=141/990` Hence, `0.bar(142)=141/990`. |
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| 21. |
Find three numbers in G.P. whose sum is 13 and the sum of whose squaresis 91. |
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Answer» Let the required numbers be `a/r, a` and `ar`. Then, `a/r+a+ar=13` ...(i) and `a^(2)/r^(2)+a^(2)+a^(2)r^(2)=91`...(ii) On squaring both sides of (i), we get `(a/r+a+ar)^(2)=169` `rArr (a^(2)/r^(2)+a^(2)+a^(2)r^(2))+2 (a^(2)/r+a^(2)+a^(2)r)=169` [using (ii)] `rArr 91+26a=169` [using (i)] `rArr 26a=78 rArr a=3`. Putting `a=3` in (i), we get `3/r+3+3r=13 rArr 3/r +3r=10 rArr 3r^(2)-10r+3=0` `rArr (r-3)(3r-1)=0rArr r=3 or r=1/3`. Hence, the required numbers are 1, 3, 9 or 9, 3, 1. |
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| 22. |
Express `0.bar(68)` as a rational number. |
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Answer» Correct Answer - `31/45` `x=0.688... rArr 10 x=6.888...` and `100x=68.888`... `:. 90 x=62 rArr x=62/90=31/45`. |
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| 23. |
Express `0.bar(123)` as a rational number. |
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Answer» Correct Answer - `123/999` `x=0.123123...` and `1000x=123.123123...` `:. 999x=123 rArr x=123/999`. |
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| 24. |
The sum of an infinite geomwtric series is 20 and the sum of the squares of these terms is 100. Find the series. |
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Answer» Correct Answer - `(8+24/5+72/25+...oo)` Let a abe the first term and r be the common ratio of the given series. Then, `a/((1-r))=20 rArr a^(2)/((1-r)^(2))=400` ...(i) And, `a^(2)/((1-r^(2)))=100` ...(ii) `:. a^(2)/((1-r)^(2))xx((1-r^(2)))/a^(2)=400/100 rArr (1+r)/(1-r)=4 rArr r=3/5` `:. a/((1-3/5))=20 rArr a=8`. Hence, the required series is `(8+24/5+72/25+...oo)`. |
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| 25. |
Prove that `6^(1//2)xx6^(1//4)xx6^(1//8)oo=6.` |
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Answer» We observe here that `(1/2+1/4+1/8+...oo)` is an infinite geometric series in which `a=1/2` and `r=(1/4xx2/1)=1/2` such that `|r| lt 1`. So, this sum is given by `S=((1/2))/((1-1/2))=((1/2))/((1/2))=1` ...(i) `:. 6^(1/2).6^(1/4).6^(1/8)...oo=6^((1/2+1/4+1/8+...oo))=6^(1)=6` [using (i)]. Hence, `6^(1/2).6^(1/4).6^(1/8)...oo=6`. |
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| 26. |
Prove that `(1-1/3+1/3^(2)-1/3^(3)+1/3^(4)-...oo)=3/4` |
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Answer» This is an infinite GP in which `a=1` and `r=(-1)/3`. `S_(oo)=a/((1-r))=1/((1+1/3))=3/4`. |
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| 27. |
Find the sum of the geometric series `3+6+12+...+1536`. |
| Answer» Correct Answer - 3069 | |
| 28. |
If `y=x+x^2+x^3+…………oo, `prove that `x= y/(1+y)` |
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Answer» By summing the given infinite GS, we get `y=x/((1-x)) rArr y(1-x)=x` `rArr y-xy=x rArr x+xy=y` `x(1+y)=yrArr x=y/((1+y))`. Hence, `x=y/((1+y))`. |
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| 29. |
If `x=2+a+a^2+oo,w h e r e|a| |
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Answer» By summing the given infinite GS, we get `x=1/((1-a))` and `y=1/((1-b))` `:. (xy)/((x+y-1))=({1/((1-a)).1/((1-b))})/({1/((1-a))+1/((1-b))-1})` `={1/((1-a)(1-b))xx((1-a)(1-b))/((1-b)+(1-a)-(1-a)(1-b))}` `=1/((1-ab))=(1-ab)^(-1)` `=(1+ab+a^(2)b^(2)+...oo)`. Hence, `(1+ab+a^(2)b^(2)+...oo)=(xy)/((x+y-1))` |
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| 30. |
How many terms of the series `2+6+18+` ... Must be taken to make the sum equal to 728 ? |
| Answer» Correct Answer - 6 | |
| 31. |
Which term of the `GP sqrt(3), 1/sqrt(3), 1/(3sqrt(3)), 1/(9sqrt(3))`, ... Is `1/(729sqrt(3))` ? |
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Answer» In the given `GP`, we have `a=sqrt(3)` and `r=(1/sqrt(3)xx1/sqrt(3))=1/3`. Let its nth term be `1/(729sqrt(3))`. Then, `T_(n)=1/(729sqrt(3))rArr ar^((n-1))=1/(729 sqrt(3))` `rArr sqrt(3)xx(1/3)^((n-1))=1/(729sqrt(3))` `rArr 1/3^((n-1))=1/((729xx3))=1/2187=1/3^(7)` `rArr n-1=7 rArr n=8`. Hence, the 8th term of the given GP is `1/(729sqrt(3))` |
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| 32. |
Find the sum of the following series:`(sqrt(2)-1)+1+(sqrt(2)-1)+oo` |
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Answer» We have `1/((sqrt(2)+1))=1/((sqrt(2)+1))xx((sqrt(2)-1))/((sqrt(2)-1))=((sqrt(2)-1))/1`. So, the given series is an infinite geometric series in which `a=(sqrt(2)+1)` and `r=(sqrt(2)-1) lt 1`. Hence, the sum of the given infinite geometric series is `S=a/((1-r))=((sqrt(2)+1))/({1-(sqrt(2)-1)})=((sqrt(2)+1))/((2-sqrt(2)))xx((2+sqrt(2)))/((2+sqrt(2)))` `=(4+3sqrt(2))/((4-2))=((4+3sqrt(2)))/2`. |
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| 33. |
For what values of x, the numbers `-2/7 , x,-7/2` are in G.P |
| Answer» Correct Answer - `x=1 or x=-1` | |
| 34. |
Which term of the G.P.: ` sqrt(3),3,3,sqrt(3), i s 729 ?` |
| Answer» Correct Answer - 12th | |
| 35. |
Find the sum to n terms of the sequence : (i) `(x+1/x)^(2)+ (x^(2)+1/x^(2))^(2)+ (x^(3)+1/x^(3))^(2)` ... to n terms(ii) `(x+y)+ (x^(2)+xy+y^(2))+(x^(3)+x^(2)y+xy^(2)+y^(3))`, ... to n terms |
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Answer» Correct Answer - (i) `((x^(2n)-1)/(x^(2)-1))(x^(2)+1/x^(2n))+2n` (ii) `1/((x-y)).{x^(2)((x^(n)-1)/(x-1))-y^(2)((y^(n)-1)/(y-1))}` (i) Given sum `=(x+1/x)^(2)+(x^(2)+1/x^(2))^(2)+(x^(3)+1/x^(3))^(2)+...+(x^(n)+1/x^(n))^(2)` `=(x^(2)+1/x^(2)+2)+(x^(4)+1/x^(4)+2)+(x^(6)+1/x^(6)+2)+...+(x^(2n)+1/x^(2n)+2)` `=(a^(2)+x^(4)+x^(6)+...+x^(2n))+(1/x^(2)+1/x^(4)+1/x^(6)+...+1/x^(2n))+(2+2+..."n imes")` `=(x^(2)[(x^(2))^(n)-1])/((x^(2)-1))+(1/x^(2).{1-(1/x^(2))^(n)})/((1-1/x^(2)))+2n=(x^(2)(x^(2n)-1))/((x^(2)-1))+((x^(2n)-1))/(x^(2n) (x^(2)-1))+2n` `=x^(2) ((x^(2n)-1)/(x^(2)-1))+1/x^(2n)((x^(2n)-1)/(x^(2)-1))+2n=((x^(2n)-1)/(x^(2)-1))(x^(2)+1/x^(2n))+2n`. (ii) Multiplying and dividing each term by `(x-y)`, we get the given sum as `1/((x-y)).{(x^(2)-y^(2))+(x^(3)-y^(3))+(x^(4)-y^(4))+..."to n terms"}` `=1/((x-y)).{(x^(2)+x^(3)+x^(4)+..."to n terms")-(y^(2)+y^(3)+y^(4)+..."to n terms")}` `=1/((x-y)).{x^(2)((x^(n)-1)/(x-1))-y^(2) ((y^(n)-1)/(y-1))}`. |
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| 36. |
Insert threenumbers between 1 and 256 so that the resulting sequence is a G.P. |
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Answer» Let the required numbers be `G_(1), G_(2), G_(3)`. Then, `1, G_(1), G_(2), G_(3), 256` are in GP. Let r be the common ratio of this GP. Then, `T_(5)=256 rArr ar^((5-1))=256` `rArr 1xxr^(4)=256 rArr r^(4) =256= (4)^(4) rArr r=4`. `:. G_(1)=(1xx4)=4, G_(2)=(1xx4^(2))=16` and `G_(3)=(1xx4^(3))=64`. Hence, the required are 4, 16, 64. |
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| 37. |
Find the 4th term from the end of the G.P. `2/(27),2/9,2/3, , 162.` |
| Answer» Correct Answer - `6` | |
| 38. |
The `6th` term from the end of the `G.P. 8, 4, 2, 1,...1/1024` (A) `1/64` (B)`32` (C)`1/32` (D)none of these |
| Answer» Correct Answer - `1/32` | |
| 39. |
Find the sum of n terms of the sequence given by `a_(n)=(3^(n)+5n), n in N`. |
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Answer» Let the sum of n terms of the given sequence be `S_(n)`. Then, `S_(n)=a_(1)+a_(2)+a_(3)+...+a_(n)` `=(3^(1)+5xx1)+(3^(2)+5xx2)+(3^(3)+5xx2)+(3^(3)+5xx3)+...+(3^(n)+5xxn)` `=(3+3^(2)+3^(3)+...+3^(n))+(5xx1+5xx2+5xx3+...+5xxn)` `=(3+3^(2)+3^(3)+...+3^(n))+5xx(1+2+3+..+n)` `=(3(3^(n)-1))/((3-1))+5xxn/2 (1+n)` `=3/2 (3^(n)-1)+5/2 xx n (n+1)`. Hence, the required sum is `3/2 (3^(n)-1)+5/2 n(n+1)`. |
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| 40. |
The sum of some terms of G. P.is 315 whose first term and the common ratio are 5 and 2, respectively. Findthe last term and the number of terms. |
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Answer» Let the given GP contain n terms, Then, `a=5, r=2 gt 1` and `S_(n)=315`. `:. S_(n)=315 rArr (a(r^(n)-1))/((r-1))=315` `rArr (5xx(2^(n)-1))/((2-1))=315` `rArr (2^(n)-1)=63 rArr 2^(n) =64=2^(6) rArr n=6`. Last term `=ar^((n-1))=5xx2^((6-1))=((5xx2^(5))=(5xx32))=160`. Hence, the given GP contains 6 terms and its last term is 160. |
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| 41. |
Find the sum of the products of the corresponding terms of finite geometrical progressions 2, 4, 8, 16, 32 and 128, 32, 8, 2, `1/2`. |
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Answer» Raking the products of the corresponding terms of two given GPs, we get a new progression `(2xx128), (4xx32), (8xx8), (16xx2), (32xx1/2)` i.e., `256, 128, 64, 32, 16`, which is clealy a GP in which `a=256` and `r=16/32=1/2 lt 1`. `:.` the required sum `=(a(1-r^(5)))/((1-r))=(256xx{1-(1/2)^(5)})/((1-1/2))` `=(256xx(1-1/2^(5)))/((1/2))=256xx2xx(1-1/32)` `=(256xx2xx31/32)=496`. |
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| 42. |
In a n increasing G.P. , the sum of the firstand the last term is 66, the product of the second and the last but one is128 and the sum of the terms is 126. How many terms are there in theprogression? |
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Answer» Let the given GP contain n terms. Let a be the first term and r be the common ratio of this GP. Since the given GP is increasing, we have `r gt 1`. Now, `T_(1)T_(n)=66 rArr a+ar^((n-1))=66`. ...(i) And, `T_(2)xxT_(n-1)=128 rArr ar xx ar^((n-2))=128` `rArr a^(2)r^((n-1))=128 rArr ar^((n-1))=128/a`. ...(ii) Using (ii) in (i), we get `a+128/a=66 rArr a^(2)-66a+128=0` `rArr a^(2)-2a-64a+128=0` `rArr a(a-2)-64(a-2)=0` `rArr (a-2)(a-64)=0` `rArr a=2 or a=64`. Putting `a=2` in (ii), we get `r^((n-1))=128/a^(2)=128/4=32`. ..(iii) Putting `a=64` in (ii), we get `r^((n-1))=128/a^(2)=128/(64xx64)=1/32`, which is rejected, sice `r gt 1`. Thus, `a=2` and `r^((n-1))=32`. Now, `S_(n)=126 rArr (a(r^(n)-1))/((r-1))=126` `rArr 2((r^(n)-1)/(r-1))=126 rArr (r^(n)-1)/(r-1)=63` `rArr (r^((n-1))xxr-1)/(r-1)=63 rArr (32r-1)/(r-1)=63` `rArr 32r-1=63 r-63 rArr 31 r=62 rArr r=2`. `:. r^((n-1))=32=2^(5) rArr n-1=5rArr n=6`. Hence, there are 6 terms in the given GP. |
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| 43. |
Which term of the G.P `1/4, -1/2, 1, ...` is `-128` |
| Answer» Correct Answer - 10th | |
| 44. |
The sum of firstthree terms of a G.P. is 16 and the sum of the next three terms is 128.Determine the first term, the common ratio and the sum to n terms of the GP. |
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Answer» Let a be the first term and r be the common ratio of the given GP. Then, `a+ar+ar^(2)=16` and `ar^(3)+ar^(4)+ar^(5)=128` `rArr a(1+r+r^(2))=16` ...(i) and `ar^(3)(1+r+r^(2))=128` ...(ii). On dividing (ii) by (i), we get `r^(3)=8=2^(3) rArr r=2`. Putting `r=2` in (i), we get `a(1+2+4)=16rArr a=16/7`. Thus, in the given GP, we have `a=16/7` and `r=2 gt 1`. `:. S_(n)=(a(r^(n)-1))/((r-1))=(16/7xx(2^(n)-1))/((2-1))=16/7 (2^(n)-1)`. |
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| 45. |
Find the sum of the infinite geometric series `(1-1/3+1/3^(2)-1/3^(3)+...oo)`. |
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Answer» The given series is an infinite geometric series in which `a=1, r=-1/3` and `|r|=1/3 lt 1` Hence, the sum of the given infinite geometric series is `S=a/((1-r))=1/((1+1/3))=1/((4/3))=3/4`. |
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| 46. |
The ratio of the sum of first three terms is tothat of first 6 terms of a G.P. is 1:12. Find the common ratio. |
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Answer» Correct Answer - `3/5` `(a(r^(3)-1))/((r-1))xx((r-1))/(a(r^(6)-1))=125/152 rArr 1/((r^(3)+1))=125/152 rArr (r^(3)+1)=152/125`. `:. r^(3)=(152/125-1)=27/125=(3/5)^(3)rArr r=3/5`. |
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| 47. |
The sum of first three terms of a G.P. is `(39)/(10)`and their product is 1. Find the common ratio and the terms. |
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Answer» Correct Answer - `r=5/2 or 2/5; 2/5, 1, 5/2 or 5/2, 1, 2/5` Let the three terms of the GP be `a/r, a` and `ar`. Then, `a/rxxaxxar=1 rArr a^(3)=1 rArr a=1`. So, the given terms are `1/r, 1` and r. `:. 1/r+1+r=39/10 rArr r+1/r=(39/10-1)=29/10` `rArr 10r^(2)+10=29r` `rArr 10 r^(2)-29r +10=0` `rArr (2r-5) (5r-2)=0` `rArr r=5/2` or `r=2/5` Hence, these terms are `2/5, 1, 5/2` or `5/2, 1, 2/5`. |
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| 48. |
Write the quadratic equation, the arithmetic and geometric means of whose roots are A and G respectively. |
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Answer» Correct Answer - `x^(2)-2Ax+G^(2)=0` Let the roots of the quadratic equation be `alpha` and `beta`. Then, the equation is `x^(2)-(alpha+beta)x+alpha beta=0`. `:. A=(alpha +beta)/2 rArr alpha+beta =2A`. `G =sqrt(alpha beta)rArr alpha beta=G^(2)`. Hence, the required equation is `x^(2)-2Ax+G^(2)=0`. |
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| 49. |
If `G_1` and `G_2` are two geometric means and A is the arithmetic mean inserted two numbers, then the value of `(G_1^2)/G_2+(G_2^2)/G_1` is: |
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Answer» Let the given numbers be a and b. Then, `A=(a+b)/2 rArr a+b=2A` ...(i) And, `a, G_(1), G_(2), b` are in GP. `:. G_(1)/a=G_(2)/G_(1)=b/G_(2)` `rArr a=G_(1)^(2)/G_(2)` and `b=G_(2)^(2)/G_(1)` `rArr a+b=G_(1)^(2)/G_(2)+G_(2)^(2)/G_(1)` `rArr 2A=G_(1)^(2)/G_(1)+G_(2)^(2)/G_(1)` [using (i)]. Hence, `2A=G_(1)^(2)/G_(2)+G_(2)^(2)/G_(1)`. |
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| 50. |
If A.M. and GM. of roots of a quadraticequation are 8 and 5, respectively, then obtain the quadratic equation. |
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Answer» Let the required roots be a and b. Then, `AM=(a+b)/2 rArr (a+b)/2=8 rArr a+b=16` ...(i) and `GM=sqrt(ab) rArr sqrt(ab)=5 rArr ab=25` ...(ii) `:.` sum of roots =16 and product of roots `=25` Hence, the required equation is `x^(2)-16x+25=0`. |
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