InterviewSolution
Saved Bookmarks
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.
| 301. |
In a G.P. of positive terms if any terms is equal to the sum of nexttwo terms, find the common ratio of the G.P.A. `(sqrt(5)-1)/(2)`B. `(sqrt(5)+1)/(2)`C. `-(sqrt(5)+1)/(2)`D. `(1-sqrt(5))/(2)` |
|
Answer» Correct Answer - A We have, Let `a_(1),a_(2),a_(3), . . . . .,a_(n), . . . .` be a G.P. with common ratio r. Then, `a_(n)=a_(n+1)+a_(n+2)` [Given] `rArr" "a_(1)r^(n-1)=a_(1)r^(n)+a_(1)r^(n+1)` `rArr" "r^(-1)=1+r` `rArrr^(2)+r-1=0rArrr=(-1pmsqrt(5))/(2)rArr=(sqrt(5-1))/(2)[becausergt0"(given)"]` |
|
| 302. |
Let `a_(1)+a_(2)+a_(3), . . . ,a_(n-1),a_(n)` be an A.P. Statement -1: `a_(1)+a_(2)+a_(3)+ . . . +a_(n)=(n)/(2)(a_(1)+a_(n))` Statement -2 `a_(k)+a_(n-k+1)=a_(1)+a_(n)" for "k=1,2,3, . . . ,` nA. Statement -1 is true, Statement -2 is True, Statement -2 is a correct explanation for Statement for Statement -1.B. Statement -1 is true, Statement -2 is True, Statement -2 is not a correct explanation for Statement for Statement -1.C. Statement -1 is true, Statement -2 is False.D. Statement -1 is False, Statement -2 is True. |
|
Answer» Correct Answer - A Let `S_(n)=a_(1)+a_(2)+a_(3)+ . . . +a_(n-1)+a_(n)`. Then, `S_(n)=a_(n)+a_(n-1)+a_(n-2)+ . . . +a_(2)+a_(1)` `rArr" "2S_(n)=n(a_(1)+a_(n))" [Using statement-2]"` `rArr" "S_(n)=(n)/(2)(a_(1)+a_(2))` So, statement -2 correct explanation for statement -1. |
|
| 303. |
Stament -1: If for any real `x,2^(1+x)+2^(1-x),lamda and3^(x)+3^(-x)` are three equidistant terms of an A.P., then `lamdage3`. Statement -2: `AMgeGM`A. Statement -1 is true, Statement -2 is True, Statement -2 is a correct explanation for Statement for Statement -1.B. Statement -1 is true, Statement -2 is True, Statement -2 is not a correct explanation for Statement for Statement -1.C. Statement -1 is true, Statement -2 is False.D. Statement -1 is False, Statement -2 is True. |
|
Answer» Correct Answer - A As `2^(1+x)+2^(1-x),lamdaand3^(x)+3^(-x)` are three equidistant terms of an A.P. `:." "2^(1+x)+2^(1-x),lamdaand3^(x)+3^(-x)` are in A.P. `rArr" "2lamda=2^(1+x)+2^(1-x)+3^(x)+3^(-x)` `rArr" "2lamda=2(2^(x)+2^(-x))+(3^(x)+3^(-x))` Now, `AMgtGM` `rArr" "2^(x)+2^(-x)ge2and3^(x)+3^(-x)ge2` `rArr" "2(2^(x)+2^(-x))+(3^(x)+3^(-x))ge4+2rArr2lamdage6rArrlamdage3` |
|
| 304. |
Suppose four distinct positive numbers `a_(1),a_(2),a_(3),a_(4)` are in G.P. Let `b_(1)=a_(1)+,a_(b)=b_(1)+a_(2),b_(3)=b_(2)+a_(3)andb_(4)=b_(3)+a_(4)`. Statement -1 : The numbers `b_(1),b_(2),b_(3),b_(4)` are neither in A.P. nor in G.P. Statement -2: The numbers `b_(1),b_(2),b_(3),b_(4)` are in H.P.A. Statement -1 is true, Statement -2 is True, Statement -2 is a correct explanation for Statement for Statement -1.B. Statement -1 is true, Statement -2 is True, Statement -2 is not a correct explanation for Statement for Statement -1.C. Statement -1 is true, Statement -2 is False.D. Statement -1 is False, Statement -2 is True. |
|
Answer» Correct Answer - C Let r be the common ratio of the G.P. We have, `b_(1)=a_(1),b_(2)=a_(1)+a_(2)=a_(1)(1+r),b_(3)=b_(2)+a_(3)=a(1+r+r^(2))`, `b_(4)=b_(3)+a_(4)=a_(1)(1+r+r^(2)+r^(3))` Clearly, `b_(1),b_(2),b_(3),b_(4)` are neither in A.P. nor in G.P. So, statement -1 is correct. Also, `b_(1),b_(2),b_(3),b_(4)` are not in H.P. So, statement -2 is false. |
|
| 305. |
Natural numbers are divided into groups in the following way: `1,(2,3),(4,5,6),(7,8,9,10),` Show that the sum of the numbers in the nth group is `(n(n^2+1)/2`A. 62525B. 65255C. 56255D. 55625 |
| Answer» Correct Answer - A | |
| 306. |
यदि= `a` तथा `b` दो अलग-अलग प्राकृत संख्याएं है तो इनमें कौन सा कथन सत्य है?A. `2sqrt(ab)gta+b`B. `2sqrt(ab)lta+b`C. `2sqrt(ab)=a+b`D. none of these |
| Answer» Correct Answer - B | |
| 307. |
If a,b,c,d are in H.P., then ab+bc+cd is equal toA. 3 adB. (a+b)(c+d)C. 3acD. none of these |
| Answer» Correct Answer - A | |
| 308. |
The sum to `n` terms of the series `3/(1^2)+5/(1^2+2^2)+7/(1^2+2^2+3^2)+-------` isA. `(6n)/(n+1)`B. `(9n)/(n+1)`C. `(12n)/(n+1)`D. `(3n)/(n+1)` |
|
Answer» Correct Answer - A Let `T_(r)` be the `r^(th)` term of the given series. Then, `T_(r)=(2r+1)/(1^(2)+2^(2)+ . . . +r^(2))=(2r+1)/((r//6)(r+1)(2r+1))=6((1)/(r)-(1)/(r+1))` So, required sum is given by `underset(r=1)overset(n)sumT_(r)=6underset(r=1)overset(n)sum((1)/(r)-(1)/(r+1))` `rArr" "underset(r=1)overset(n)sum*T_(r)=6{(1-(1)/(2))+((1)/(2)-(1)/(3))+((1)/(3)-(1)/(4))+ . . . +((1)/(n)-(1)/(n+1))}` `rArr" "underset(r=1)overset(n)sumT_(r)=6{1-(1)/(n+1)}=(6n)/(n+1)` |
|
| 309. |
The sum of `n` terms of the series `1/(sqrt1+sqrt3)+1/(sqrt3+sqrt5)+...` isA. `sqrt(2n+1)`B. `(1)/(2)sqrt(2n+1)`C. `sqrt(2n+1)-1`D. `(1)/(2)(sqrt(2n+1)-1)` |
|
Answer» Correct Answer - D The required sum to n terms is given by `(1)/(sqrt(1)+sqrt(3))+(1)/(sqrt(3)+sqrt(5))+(1)/(sqrt(5)+sqrt(7))+ . . .+(1)/(sqrt(2n-1)+sqrt(2n+1))` `=(1)/(2)[(sqrt(3)-sqrt(1))+(sqrt(5)-sqrt(3))+(sqrt(7)-sqrt(5))+(sqrt(2n+1)-sqrt(2n-1))]` `=(1)/(2)(sqrt(2n+1)-1)` |
|
| 310. |
The `7th` term of an `H.P.` is `1/10` and `12th` term is `1/25` Find the `20th` termA. `(1)/(37)`B. `(1)/(41)`C. `(1)/(45)`D. `(1)/(49)` |
| Answer» Correct Answer - C | |
| 311. |
`1/a+1/c+1/(a-b)+1/(c-b)=0` and `b!=a+c,` then `a,b,c ` are inA. H.P.B. G.P.C. A.P.D. none of these |
| Answer» Correct Answer - B | |
| 312. |
If `S_n=1/1^3 +(1+2)/(1^3+2^3)+...+(1+2+3+...+n)/(1^3+2^3+3^3+...+n^3)` Then `S_n` is not greater thanA. `(1)/(2)`B. 1C. 2D. 4 |
| Answer» Correct Answer - C | |
| 313. |
The harmonic mean of the roots of the equation `(5+sqrt(2))x^2-(4+sqrt(5))x+8+2sqrt(5)=0`is`2`b. `4`c. `6`d. `8`A. 2B. 4C. 6D. 8 |
| Answer» Correct Answer - D | |
| 314. |
Find the harmonic mean of the roots of the equation `(5+sqrt2)x^2-(4+sqrt5)x+(8+2sqrt5)=0`A. 2B. 4C. 6D. 8 |
|
Answer» Correct Answer - B Let `alpha, beta` be the roots of given quadratic equation. Then, `alpha + beta = (4 + sqrt5)/(5 + sqrt2) and alpha beta = (8 + 2 sqrt5)/(5 + sqrt2)` Let H be the harmonic mean between `alpha and beta`, then `H = (2 alpha beta)/(alpha + beta) = (16 + 4 sqrt5)/(4 + sqrt5) = 4` |
|
| 315. |
If the geometric mea is `(1)/(n)` times the harmonic mean between two numbers, then show that the ratio of the two numbers is `1+sqrt(1-n^(2)):1-sqrt(1-n^(2))`. |
|
Answer» Let the two numbers be a and b. Given, `G=(1)/(n)H " " "….(i)"` Now, `G^(2)=AH` `implies (H^(2))/(n^(2))=AH " " [ " From Eq. (i) " ]` `therefore A=(H)/(n^(2))" " "…..(ii)"` Now, from important theorem of GM `a,b=Apm sqrt((A^(2)-G^(2)))=(H)/(n^(2))pm sqrt(((H^(2))/(n^(4))-(H^(2))/(n^(2))))` `(H)/(n^(2))[1pm sqrt((1-n^(2)))]` `therefore (a)/(b)=((H)/(n^(2))[1+ sqrt((1-n^(2)))])/((H)/(n^(2))[1- sqrt((1-n^(2)))])` `therefore a:b=1+sqrt((1-n^(2))):1-sqrt((1-n^(2)))` |
|
| 316. |
The sum of the first 20 terms of the series `1+3/2+7/4+15/8+31/16+...` is:A. `2(n-1)+(1)/(2n-1)`B. `2n-(1)/(2^(n))`C. `2+(1)/(2^(n))`D. `2n-1+(1)/(2^(n))` |
| Answer» Correct Answer - A | |
| 317. |
The sum of the first n terms of the series `(1)/(2)+(3)/(4)+(7)/(8)+(15)/(16)+....` is equal toA. `2^(n)-n-1`B. `1-2^(-n)`C. `n+26(-n)-1`D. `26(n)-1` |
| Answer» Correct Answer - B | |
| 318. |
If the first four terms of an arithmetic sequence are `a,2a,b` and `(a-6-b)` for some numbers a and b, find the sum of the first 100 terms of the sequence. |
|
Answer» First four terms of an AP are a, 2a ,b and (a-6-b) So ` " " 2a - a=a-6-b-b` ` rArr " " a= a-6-2b ` `rArr " " - 2b =6rArr b=-3 ` ` and " " 2a-a = b -2a ` ` rArr " " b= 3a rArr a=-1 ` ` therefore ` First terms ` a=-1 and d=a-1 ` ` S_(100) =(100)/( 2) [2a +(100-1)d]` ` " "= 50[-2+99(-1) ]` ` " " 50 (-2-99)= 50)(-101)=- 5050`. |
|
| 319. |
The sequence of odd natural numbers is divided into groups `1,3,5,7,9,11,"…."` and so on. Show that the sum of the numbers in nth group is `n^(3)`. |
|
Answer» Sequence of natural number is divided into group 1,3,5,7,9,11… ` therefore ` nth row contains n elements 1st elements of nth row =`n^(2) -( n-1) ` Least elements of nth row ` =n^(2) +( n-1) ` ` therefore `Sum of the element in the nth row ltbr gt ` = (n)/(2) (a+l) =(n)/(2) [n^(2) + n-1] =(n)/(2) [2n ^(2) ] ]=n^(3) ` ` =(n)/(2) [n^(2) -n+1+ n^(2) + n-1])(n)/(2) [2n^(2) ] =n^(3)`. |
|
| 320. |
If the ratio of `H.M.` and `G.M.` between two numbers `a` and `b` is `4:5`, then find the ratio of the two number ?A. `4:1`B. `3:2`C. `3:4`D. `2:3` |
|
Answer» Correct Answer - A Let the numbers be a and b. Then, `((2ab)/(a+b))/(sqrt(ab))=(4)/(5)` `rArr" "(2sqrt(ab))/(a+b)=(4)/(5)` `rArr" "(a+b)/(2sqrt(ab))=(5)/(4)` `rArr" "(a+b+2sqrt(ab))/(a+b-2sqrt(ab))=(5+4)/(5-4)" "{:[("Applying componendo"),("and dividendo")]:}` `rArr" "((sqrt(a)+sqrt(b))/(sqrt(a)-sqrt(b)))^(2)=(9)/(1)` `rArr(sqrt(a)+sqrt(b))/(sqrt(a)-sqrt(b))=(3)/(1)rArr(sqrt(a))/(sqrt(b))=(3+1)/(3-1)rArr(a)/(b)=(4)/(1)` |
|
| 321. |
The sum to infinity of the series `1+(4)/(5)+(7)/(5^(2))+(10)/(5^(3))+ . . . ,` isA. `(16)/(35)`B. `(11)/(8)`C. `(35)/(16)`D. `(8)/(6)` |
| Answer» Correct Answer - C | |
| 322. |
Find the sum upto n terms of the series `1*4*7*10*13*16+"...."`. |
|
Answer» Let `T_(n)` be the nth term of the given series. ` therefore T_(n)=(" nth term of " 1,4,7, "..."(" nth term of " 4,7,10,"......")` `(" nth term of " 7,10,13, "...")(" nth term of " 10,13,16,"......")` `T_(n)=(3n-2)(3n+1)(3n+4)(3n+7)" " ".........(i)"` ` therefore V_(n)=(3n-2)(3n+1)(3n+4)(3n+7)(3n+10)` ` V_(n-1)=(3n-5)(3n-2)(3n+1)(3n+4)(3n+7)` `implies V_(n)=(3n+10)T_(n) " " [" from Eq. (i) "]` and ` V_(n-1)=(3n-5)T_(n)` ` therefore V_(n)-V_(n-1)=15T_(n)` `therefore T_(n)=(1)/(15)(V_(n)-V_(n-1))` `therefore S_(n)=sumT_(n)=sum_(n=1)^(n)(1)/(15)(V_(n)-V_(n-1))` `=(1)/(15)(V_(n)-V_(0)) " " [" from important Theorem 1 of " sum]` `=(1)/(15){(3n-2)(3n+1)(3n+4)(3n+7)(3n+10)-(-2)(1)(4)(7)(10)}` `=(1)/(15){(3n-2)(3n+1)(3n+4)(3n+7)(3n+10)+560}` Shortcut Method `S_(n)=(1)/(" last factor of III term - first factor of I term ")` ( Taking one extra factor in `T_(n)` in last - Taking one exra factor in I term in start ) `=(1)/((16-1)){(3n-2)(3n+1)(3n+4)(3n+7)(3n+10)-(-2)*1*4*7*10}` `=(1)/(15){(3n-2)(3n+1)(3n+4)(3n+7)(3n+10)+560}` |
|
| 323. |
Let `A_1 , G_1, H_1`denote the arithmetic, geometric and harmonic means respectively, of two distinct positive numbers. For `n >2,`let `A_(n-1),G_(n-1)` and `H_(n-1)` has arithmetic, geometric and harmonic means as `A_n, G_N, H_N,` respectively.A. `G_(1)gtG_(2)gtG_(3)gt . . .`B. `G_(1)ltG_(2)ltG_(3)lt . . .`C. `G_(1)=G_(2)=G_(3)= . . .`D. `G_(1)ltG_(3)ltG_(5)= . . .andG_(2)gtG_(4)gtG_(6)gt . . .` |
|
Answer» Correct Answer - C Let a and b be two distinct positive numbers. Then, `A_(1)=(a+b)/(2),G_(1)sqrt(ab)andH_(1)=(2ab)/(a+b)` It is given that for `nge2` `A_(n)=(A_(n-1)+H_(n-1))/(2),G_(n)=sqrt(A_(n-1)H_(n-1))` `andH_(n)=(2A_(n-1)H_(n-1))/(A_(n-1)+H_(n-1))` `rArr" "A_(n)H_(n)=A_(n-1)H_(n-1)` `:." "G_(n+1)=sqrt(A_(n)H_(n))=sqrt(A_(n-1)H_(n-1))=G_(n)"for "nge2` `rArr" "G_(n)=G_(n-1)=G_(n-2)= . . .=G_(3)=G_(2)=G_(1)=sqrt(ab)`. |
|
| 324. |
The sum to infinity of the series `1+2(1-(1)/(n))+3(1-(1)/(n))^(2)+ . . .. . . .`, isA. `n^(2)`B. `n(n+1)`C. `n(1+(1)/(n))^(2)`D. None of these |
|
Answer» Correct Answer - A Let `S=1+2(1-(1)/(n))+3(1-(1)/(n))^(2)+"...."+oo` `((1-(1)/(n))S=(1-(1)/(n))+2(1-(1)/(n))+"...."+oo)/(S(1-1+(1)/(n))=1(1-(1)/(n))+(1-(1)/(n))^(2)+"...."+oo)` `implies (S)/(n)=(1)/(1-(1-(1)/(n))) " " [S_(oo)=(a)/(1-r) " by GP "]` `implies S=(n)/((1)/(n))` `implies S=n^(2)`. |
|
| 325. |
Sum of the following series to `n`term: `2+4+7+11+16+ ` |
|
Answer» The sequence of first consctive differences is `2,3,4,5,"…"`. Clearly, it is an AP. Then, nth term of the given series be `T_(n)=a(n-1)(n-2)+b(n-1)+c " " "....(i)"` Putting `n=1,2,3,` we get `2=c implies 4=b+c implies 7=2a+2b+c` After solving, we get `a=(1)/(2),b=2,c=2` Putting the values of `a,b,c` in Eq. (i), we get `T_(n)=(1)/(2)(n-1)(n-2)+2(n-1)+2=(1)/(2)(n^(2)+n+2)` Hence, sum of series `S_(n)=sum T_(n)=(1)/(2)(sumn^(2)+sumn+2sum1)` `=(1)/(2)((n(n+1)(2n+1))/(6)+(n(n+1))/(2)+2n)` `=(1)/(6)n(n^(2)+3n+8)` |
|
| 326. |
Find the nth term and sum to n terms of the series `12+40+90+168+280+432+"...."`. |
|
Answer» Let `S_(n)=12+40+90+168+280+432+"....",` then 1st order differences are `28,50,78,112,152,"...."(i.e.Deltat_(1),Deltat_(2),Deltat_(3),".....")` and 2nd order differences are `22,28,34,40,"...."(i.e.Deltat_(1)^(2),Deltat_(2)^(2),Deltat_(3)^(2),".....")` and 3rd order dofferences are `6,6,6,6,"...."(i.e.Deltat_(1)^(3),Deltat_(2)^(3),Deltat_(3)^(3),".....")` and 4th order differences are `0,0,0,0,"...."(i.e.Deltat_(1)^(4),Deltat_(2)^(4),Deltat_(3)^(4),".....")` `therefore t_(n)=12*^(n-1)C_(0)+28*^(n-1)C_(1)+22*^(n-1)C_(2)+6*^(n-1)C_(3)` `=12+28(n-1)+(22(n-1)(n-2))/(2)+(6(n-1)(n-2)(n-3))/(1*2*3)` `=n^(3)+5n^(2)+6n` and `S_(n)=12*^(n)C_(1)+28*^(n)C_(2)+22*^(n)C_(3)+6*^(n)C_(4)` `=12n+(28n(n-1))/(2)+(22n(n-1)(n-2))/(1*2*3)+(6*n(n-1)(n-2)(n-3))/(1*2*3*4)` `=(n)/(12)(n+1)(3n^(2)+23n+46)`. |
|
| 327. |
In Illustration 6, which one of the following statement is correct ?A. `A_(1)gtA_(2)gtA_(3)gt . . .`B. `A_(1)ltA_(2)ltA_(3) . . .`C. `A_(1)gtA_(3)gtA_(5)gt. . .andA_(2)ltA_(4)ltA_(6)lt`D. `A_(1)ltA_(3)ltA_(5) lt. . .andA_(2)gtA_(4)gtA_(6)gt . . .` |
|
Answer» Correct Answer - A We have, `A_(1)gtG_(1)gtH_(1)` It is given that `A_(2)andH_(1)andA_(1)gtH_(1)`. `:." "A_(1)gtA_(2)gtH_(1)" "{:[(because"Arithmetic mean lies"),("between the numbers")]:}` `A_(3)` is the A.M. of `A_(2)andH_(2)andA_(2)gtH_(2)`. `:." "A_(2)gtA_(3)gtH_(2)" "{:[(because"Arithmetic mean lies"),("between the numbers")]:}` Hence, `A_(1)gtA_(2)gtA_(3)gt . . . . . .` |
|
| 328. |
`7^(2)+8^(2)+9^(2)+...+20^(2)=`A. 2779B. 7279C. 7729D. none of these |
| Answer» Correct Answer - A | |
| 329. |
`1^(3)+2^(3)+3^(3)+ cdots +20^(3)`=A. 14400B. 44100C. 41400D. none of these |
| Answer» Correct Answer - B | |
| 330. |
The sum of the 10 terms of the series `(x+1/x)^2+(x^2+1/x^2)^2+(x^3+1/x^3)^2+...` isA. `((x^(20)-1)/(x^(2)-1))((x^(22)+1)/(x^(20)))+20`B. `((x^(18)-1)/(x^(2)-1))((x^(11)+1)/(x^(9)))+20`C. `((x^(18)-1)/(x^(2)-1))((x^(11)-1)/(x^(9)))+20`D. none of these |
| Answer» Correct Answer - A | |
| 331. |
Find the sum ofintegers from 1 to 100 that are divisible by 2 or 5.A. 3000B. 3010C. 3150D. 3050 |
| Answer» Correct Answer - D | |
| 332. |
If `(a^n+b^n)/(a^(n-1)+b^(n-1))` is the AM between a and b, then the value of n is |
|
Answer» Correct Answer - B Since `(a^(n)+b^(n))/(a^(n-1)+b^(n-1))` is Am of a and b. `:." "(a^(n)+b^(n))/(a^(n-1)+b^(n-1))=(a+b)/(2)` `rArr" "2a^(n)+2b^(n)=a^(n)+b^(n)+ab^(n-1)+a^(n-1)b` `rArr" "a^(n-1)(a-b)=b^(n-1)(a-b)` `rArr" "((a)/(b))^(n-1)=1rArrn-1=0rArrn=1` ALITER We have, `(a^(n)+b^(n))/(a^(n-1)+b^(n-1))(a+b)/(2)` `rArr" "(a^(n)+b^(n))/(a^(n-1)+b^(n-1))=(a+b)/(2)=(a^(1)+b^(1))/(a^(0)+b^(0))` On comparing the two sides, we obtain n=0. |
|
| 333. |
The arithmetic mean between two numbers is A and the geometric mean is G. Then these numbers are:A. S=nAB. A=nSC. A=SD. none of these |
|
Answer» Correct Answer - A Let x and be two numbers. It is given that `A=(x+y)/(2)andS=n((x+y)/(2))rArrS=nA` |
|
| 334. |
The third term of a geometric progression is 4. The production of thefirst five terms isA. `4^(3)`B. `4^(5)`C. `4^(4)`D. none of these |
|
Answer» Correct Answer - B Let a be the first term and r be the common ratio. We have, `a_(3)=4rArrar^(2)=4` `:.` Product of first terms = a ar `ar^(2)ar^(3)ar^(4)=(ar^(2))=(ar^(2))^(5)=4^(5)` |
|
| 335. |
(ii)If the third term of G.P.is `4`, then find the product of first five termsA. `4^(3)`B. `4^(4)`C. `4^(5)`D. `4^(6)` |
| Answer» Correct Answer - C | |
| 336. |
If `|{:(a,b,aalpha-b),(b,c,balpha-c),(2,1," "0):}|=0andalpha!=1//2`, then a,b,c are inA. A.P.B. G.P.C. H.P.D. none of these |
|
Answer» Correct Answer - B Applying `C_(3)toC_(3)-alphaC_(1)+C_(1)+C_(2)` to the given determinant, we get `|{:(a,b," "0),(b,c," "0),(2,1,-2alpha+1):}|=0` `rArr" "(1-2alpha)(ac-b^(2))=0` `rArr" "ac=b^(2)" "[becausealpha!=1//2]` `rArr" "` a,b,c are in G.P. |
|
| 337. |
If `tan^-1x, tan^-1y` and `tan^-1z` are in A.P. then find the algebraic relation between x,y and z. If `x,y,z` are also in A.P. then show that `x=y=z` and `y!=0`A. `x=y=zory!=1`B. `x=1//z`C. x=y=z, but their common value is not necessarily zeroD. x=y=z=0 |
|
Answer» Correct Answer - C `:." "y^(2)=xz` . . .(i) Also, `tan^(-1)x,tan^(-1)y,tan^(-1)z` are in A.P. `:." "2tan^(-1)y,tan^(-1)x,tan^(-1)z` `rArr" "tan^(-1)((2y)/(a-y^(2)))=tan^(-1)((x+z)/(1-xz))` `rArr" "(2y)/(1-y^(2))=(x+z)/(1-xz)` `rArr" "(2y)/(1-xz)=(x+z)/(1-xz)" [Using (i)]"` `rArr" "` x,y,z are in A.P. . . . (ii) From (i) and (ii), we get x=y=z |
|
| 338. |
If `tan^-1x, tan^-1y` and `tan^-1z` are in A.P. then find the algebraic relation between x,y and z. If `x,y,z` are also in A.P. then show that `x=y=z` and `y!=0`A. x=y=zB. xy=yzC. `x^(2)=yz`D. `z^(2)=xy` |
|
Answer» Correct Answer - A We have, x,y,z are in A.P. `rArr" "2y=x+z` . . .(i) Also, `tan^(-1)xtan^(-1)z` are in A.P. `rArr" "2tan^(-1)y=tan^(-1)x+tan^(-1)z` `rArr" "tan^(-1)((2y)/(1-y^(2)))=tan^(-1)((x+z)/(1-xz))` `rArr" "(2y)/(1-y^(2))=(x+z)/(1-xz)` `rArr" "1-y^(2)=1-xz" [Using (i)]"` `rArr" "y^(2)=xz` `rArr" "` x,y,z are in G.P. . . . (ii) From (i) and (ii), we get, x=y=z |
|
| 339. |
Insert 4 A.M.s between 4 and 19. |
|
Answer» `a_n=a+(n-1)d` `119=4+(6-1)d` `d=3` 4,7,10,13,16,19. |
|
| 340. |
If x> 0, then the expression `x^100/(1 + x + x^2 + x^3 .....x^200)` is always less than or equal to |
|
Answer» `AM>=GM` `(1+x+x^2+x^3+x^4...X^200)/201>=(11*x^2*x^3...x^200)^(1/201)` `1/201>=(x^((200+201)/2))^(1/201)/(1+x+x^2+...+x^200)` `1/201>=x^100/(1+x+x^2+...+x^200)` option 1 is correct. |
|
| 341. |
Find the 23rd term of the A.P. 7,5,3,1,... |
|
Answer» In the given AP, First term, `a = 7` Common difference , `d = -2` `n_(th)` term of AP ` = a+(n-1)d` `:. 23^(rd)` term of the given AP `=7+(23-1)(-2) = 7-44 = -37.` |
|
| 342. |
The sum of the series `1 +3/4+7/16 +15/64+31/256+....` to infinite is: |
|
Answer» `=4sum_(n=1)^oo 2^n/4^n-4sum_(n=1)^oo1/4^n` `4sum_(n=1)^oo 1/2^n-4sum_(n=1)^oo 1/4^n` `4[sum_(n=1)^oo 1/2^n-sum_(n=1)^oo 1/4^n]` `4[(1/2)/(1-1/2)-(1/4)/(1-1/4)]` `4[(1/2)/(1/2)-(1/4)/(3/4)]` `4[1-1/3]=4*2/3=8/3`. |
|
| 343. |
If `log_3 2, log_3 (2^x -5)` and `log_3 (2^x-7/2)` are in `A.P`, determine the value of `x`.A. 2B. 3C. 4D. `2,3` |
|
Answer» Correct Answer - B `:. log_(3)2,log_(3)(2^(x)-5)` and `log_(3)(2^(x)-(7)/(2))" are in AP. " " " "……..(i)"` For defined, `2^(x)-5gt0` and `2^(x)-(7)/(2)gt0` `:. 2^(x)gt5 " " "…(ii)"` From Eq.(i),`2,2^(x)-5,2^(x)-(7)/(2)` are in GP. `:." " (2^(x)-5)^(2)=2*(2^(x)-(7)/(2))` `implies 2^(2x)-12*2^(x)+32=0` `implies (2^(x)-8)(2^(x)-4)=0` `:." " 2^(x)=8,4` `implies 2^(x)=8=2^(3),2^((x) ne4 " " [" from Eq. (ii) "]`. |
|
| 344. |
`S_(n)` be the sum of n terms of the series `(8)/(5)+(16)/(65)+(24)/(325)+"......"` The seveth term of the series isA. `(56)/(2505)`B. `(56)/(6505)`C. `(56)/(5185)`D. `(107)/(9605)` |
|
Answer» Correct Answer - D Let `S_n=(8)/(5)+(16)/(65)+(24)/(325)+"......"` `T_(r)=(8r)/(4r^(4)+1)=(8r)/((2^(2)+2r+1)(2^(2)-2r+1))` `=2[((2^(2)+2r+1)-(2^(2)-2r+1))/((2^(2)+2r+1)(2^(2)-2r+1))]` `=2[(1)/((2^(2)-2r+1))-(1)/((2^(2)+2r+1))]` `T_(7)=(8xx7)/(4xx7^(2)+1)=(56)/(9605)`. |
|
| 345. |
`S_(n)` be the sum of n terms of the series `(8)/(5)+(16)/(65)+(24)/(325)+"......"` The value of `S_(8)`, isA. `(288)/(145)`B. `(1088)/(545)`C. `(81)/(41)`D. `(107)/(245)` |
|
Answer» Correct Answer - A Let `S_n=(8)/(5)+(16)/(65)+(24)/(325)+"......"` `T_(r)=(8r)/(4r^(4)+1)=(8r)/((2^(2)+2r+1)(2^(2)-2r+1))` `=2[((2^(2)+2r+1)-(2^(2)-2r+1))/((2^(2)+2r+1)(2^(2)-2r+1))]` `=2[(1)/((2^(2)-2r+1))-(1)/((2^(2)+2r+1))]` `S_(8)=sum_(r=1)^(8)T_(r)=2sum_(r=1)^(8)((1)/(2^(2)-2r+1)-(1)/(2^(2)+2r+1))` `=2(1-(1)/(2(8)^(2)+2(8)+1))=2(1-(1)/(145))=(288)/(145)` |
|
| 346. |
If p is the first of the n arithmetic means between two numbers and q be the first on n harmonic means between the same numbers. Then, show that q does not lie between p and `((n+1)/(n-1))^2 p.`A. q lies between p and `((n+1)/(n-1))^(2)p`B. q lies between p and `((n+1)/(n-1))p`C. q does not lie between p and `((n+1)/(n-1))^(2)p`D. q does not lie between p and `((n+1)/(n-1))p` |
|
Answer» Correct Answer - C For `ngt1`, we have `n+1gtn-1` `implies (n+1)/(n-1)gt1 implies p((n+1)/(n-1))^(2)gtp" " [:.pgt0]".......(i)"` Now, `p=a+d` Since,a,p,b are in AP. And `d=(b-a)/(n+1)` Again, ` (p)/(q)=((an+b)/(n+1))xx[(a+bn)/(ab(n+1))]` ` =(a^(2)n+abn^(2)+b^(2)n+ab)/(ab(n+1)^(2))` ` =(n((a)/(b)+(b)/(a))+(n^(2)+1))/((n+1)^(2))` `implies (p)/(q)-1=(n((a)/(b)+(b)/(a)-2))/((n+1)^(2))=(n(sqrt(a)/sqrt(b)+sqrt(b)/sqrt(a))^(2))/((n+1)^(2))` So, `(p)/(q)-1gt0 implies (p)/(q)gt1 implies pgtq" " "........(iii)"` From Eqs. (i) and (ii), we get `qltplt((n+1)/(n-1))^(2)p`. |
|
| 347. |
Suppose p is the first of `n(ngt1)` arithmetic means between two positive numbers a and b and q the first of n harmonic means between the same two numbers. The value of p isA. `(na+b)/(n+1)`B. `(nb+a)/(n+1)`C. `(na-b)/(n+1)`D. `(nb-a)/(n+1)` |
|
Answer» Correct Answer - A For `ngt1`, we have `n+1gtn-1` `implies (n+1)/(n-1)gt1 implies p((n+1)/(n-1))^(2)gtp" " [:.pgt0]".......(i)"` Now, `p=a+d` Since,a,p,b are in AP. And `d=(b-a)/(n+1)` `p=a+((b-a))/(n+1)=(na+b)/(n+1)`. |
|
| 348. |
Suppose p is the first of `n(ngt1)` arithmetic means between two positive numbers a and b and q the first of n harmonic means between the same two numbers. The value of q isA. `((n-1)ab)/(nb+a)`B. `((n+1)ab)/(nb+a)`C. `((n-1)ab)/(na+b)`D. `((n-1)ab)/(na+b)` |
|
Answer» Correct Answer - B For `ngt1`, we have `n+1gtn-1` `implies (n+1)/(n-1)gt1 implies p((n+1)/(n-1))^(2)gtp" " [:.pgt0]".......(i)"` Now, `p=a+d` Since,a,p,b are in AP. And `d=(b-a)/(n+1)` `(1)/(q)=(1)/(a)+D=(1)/(a)+((1)/(b)-(1)/(a))/(n+1)` `implies q=(ab(n+1))/(a+bn)`. |
|
| 349. |
let `alpha,beta` be the roots of `x^2-x+p=0` and `gamma,delta` be the roots of `x^2-4x+q=0` . if `alpha,beta,gamma,delta` are in `G.P.` then the integral values of `p&q` respectively , are |
|
Answer» given that eqn `x^2- x + p = 0` having roots `(alpha, beta)` `x^2 - 4*x +q = 0` having roots `(gamma, delta)` `alpha, beta, gamma, delta` are in GP terms will be as `a, a*r , ar^2, a*r^3` in GP `alpha + beta= 1 & gamma + delta = 4` `alpha*beta= p & gamma*delta= q` now, `a + a*r = 1` `a(1+r) = 1` eqn (1) `a^2*r = p` eqn (2) `a*r^2 + a*r^3 = 4` `a*r^2 *(1+r) = 4` eqn (3) `a^2*r^5= q` eqn (4) dividing eqn 3 by 1` (a*r^2*(r+1))/(a*(r+1)) = 4/1` `r^2 = 4` `r = +- 2` now, ` a(r+1) = 1` ` r= 2, a=1/3` `a= 1/(r+1) , r=-2, a=-1` `alpha= -1 , beta= 2, gamma = -4 , delta = 8` `alpha = 1/3 , beta = 2/3 , gamma = 4/3 , delta = 8/3` `p = alpha* beta = a*a*r = -2` `q= gamma* delta = -4*8 =-32` |
|
| 350. |
There are two sets A and B each of which consists of three numbers in GP whose product is 64 and R and r are the common ratios sich that `R=r+2`. If `(p)/(q)=(3)/(2)`, where p and q are sum of numbers taken two at a time respectively in the two sets. The value of p isA. 66B. 72C. 78D. 84 |
|
Answer» Correct Answer - D Let `A={(A)/(R),A,AR}` `B={(a)/(r ),a,ar}` According to the question, `(A)/(R )*A*AR=64` `implies A^(3)=64 " "implies A=4" " "……..(i)"` `(a)/(r )*a*ar=64" " implies a^(3)=64 " " implies a=4 " " "……(ii)"` and `R=r+2" " "……..(iii)"` `p=(A)/(R)*A*AR+AR*(A)/(R)` `=(A^(2))/(R)+A^(2)R+A^(2)=(16)/(R )+16R+16` `q=(a)/(r )*a+a*ar+ar*(a)/( r)` `=(a^(2))/(r )+a^(2)r+a(^2)=(16)/(r )=(16)/(r )+16r+16` Give that, `(p)/(q)=(3)/(2)` So, `((16+16R^(2)+16R)r)/((16+16r^(2)+16r)R)=(3)/(2)` `((1+R^(2)+R)r)/((1+r^(2)+r)R)=(3)/(2)` From Eq. (iii), `R=r+2` `implies ((1+r^(2)+4+4r+r+2)r)/((1+r+r^(2))(r+2))=(3)/(2)` `implies (r^(3)+5r^(2)+7r)/(r^(3)+3r^(2)+3r+2)=(3)/(2)` `implies r^(3)-r^(2)-5r+6=0` `implies (r-2)(r^(2)+r-3)=0` `impliesr=2" or " r=(-1pmsqrt(13))/(2)` So, `R=4` `p=16((1)/(R)+R+1)=16((1)/(4)+4+1)=(16)/(4)(21)=84` |
|