SEQUENCE AND SERIES QUIZ-1

SEQUENCE AND SERIES QUIZ-1

Dear Readers,

As per analysis for previous years, it has been observed that students preparing for JEE MAINS find Mathematics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Mathematics due to their ENGINEERING background. 

Furthermore, sections such as Mathematics are dominantly based on theories, laws, numerical in comparison to a section of Engineering which is more of fact-based, Physics, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced.

Q1. The sum of the series 1+3x+6x^2+10x^3+...∞ will be

•  1/(1-x)^2
•  1/(1-x)
•  1/(1+x)^2
•  1/(1-x)^3

Solution
Let S=1+3x+6x^2+10x^3+...∞ …(i) xS=x+3x^2+6x^3+...∞ …(ii) On subtracting Eq. (ii) from Eq. (i), we get S(1-x)=1+2x+3x^2+4x^3+...∞ …(iii) ⇒ x(1-x)S=x+2x^2+3x^3+...∞ …(iv) Again, subtracting Eq. (iv) from Eq. (iii), we get S[(1-x)-x(1-x)]=(1+x+x^2+x^3+...∞) ⇒S[(1-x)(1-x)]=1/(1-x) ⇒S=1/(1-x)^3

Q2.If 0

•  log⁡(y^3/(2-y^3 ))
•  log⁡(y^3/(1-y^3 ))
•  log⁡((2y^3)/(1-y^3 ))
•   log⁡(y^3/(1-2y^3 ))

Solution
x(y^3-1)=1⇒x=1/(y^3-1)=1/k [say] ⇒k=1/x Then, 2/x+2/(3x^3 )+2/(5x^5 )+...=2k+2/3 k^3+2/5 k^5+... =log⁡〖(1+k)/(1-k)〗=log⁡{(1+y^3-1)/(1-y^3+1)} =log⁡〖y^3/(2-y^3 )〗

Q3.  If (3+5+7+⋯+n terms)/(5+8+11+⋯+10 terms)=7, then the value of n is

•   35
•  36
•  37
•  40

Solution
We have, (3+5+7+⋯+n (terms))/(5+8+11+⋯+10 terms)=7 ⇒(n/2 {6+(n-1)2})/(10/2 {10+(10-1)3} )=7 ⇒(n(n+2))/(5×37)=7 ⇒n^2+2n=35×37⇒(n+37)(n-35)=0⇒n=35

Q4. If S is the sum of an infinite GP, the first term a, then the common ratio r is given by

•  (a-S)/S
•  (S-a)/S
•  a/(1-S)
•  (S-a)/a

Solution
Sum of an infinite GP = a/(1-r)=S ⇒ a=S(1-r) ⇒ r=(S-a)/S

Q5.If 2 log_8⁡〖a=x,log_2⁡〖2a=y〗 〗 and y-x=4, then x=

•  10
•  16
•  4
•  6

Solution
 We have, 2 log_8⁡〖a=x,log_2⁡〖2a=y〗 〗 and y-x=4 ⇒2/3 log_2⁡〖a=x 〗and log_2⁡〖2+log_2⁡〖a=y〗 〗 and y-x=4 ⇒1+3/2 x=y and y-x=4⇒x=6

Q6. If log⁡〖(1-x+x^2 )=a_1 x^2+a_3 x^3 〗+⋯, then a_3+a_6+a_9+⋯ is equal to

•  log⁡〖 2〗
•   2/3 log⁡〖 2〗
•  1/3 log⁡〖 2〗
•  2 log⁡〖 2〗

Solution
We have a_n=(2(-1)^(n-1))/n, if n is a multiple of 3 ∴a_3+a_6+a_9+⋯ =2(1/3-1/6+1/9-1/12+⋯) =2/3 (1-1/2+1/3-1/4+⋯)=2/3 log⁡2 ALITER We have log⁡〖(1-x+x^2)〗=x(a_1+a_4 x^3+a_7 x^6+⋯)+x^2 (a_2+a_5 x^3+⋯) +(a_3 x^3+a_6 x^6+⋯) Put x=1,ω,ω^2 respectively and add to get the value of a_3+a_6+a_9+⋯

Q7.If log⁡3/(x-y)=log⁡5/(y-z)=log⁡7/(z-x), then 3^(x+y) 5^(y+z) 7^(z+x)=

•  0
•  2
•  1
•  None Of these

Solution
We have, log⁡3/(x-y)=log⁡5/(y-z)=log⁡7/(z-x)=λ(say) ⇒log⁡3=λ(x-y),log⁡〖5=λ(y-z)〗,log⁡〖7=λ(z-x)〗 ⇒3=10^(λ(x-y)),5=10^(λ(y-z)),7=10^(λ(z-x)) ⇒3^(x+y).5^(z+x).7^(z+x)=10^λ(x^2-y^2 ) .10^λ(y^2-z^2 ) .10^λ(z^2-x^2 ) ⇒3^(x+y).5^(y+z).7^(z+x)=10^(λ(x^2-y^2+y^2-z^2+z^2-x^2))=10^0=1

Q8.If a_1,a_2,a_3,…,a_n are in A.P. and a_i>0 for each i=1,2,3,…,n, then ∑_(r=1)^(n-1)▒1/(a_(r+1)^(2/3)+a_(r+1)^(1/3) 〖 a〗_r^(1/3)+a_r^(2/3) ) is equal to

•  (n+1)/(a_(n-1)^(2/3)+a_(n-1)^(1/3) a_1^(1/3)+a_1^(2/3) )
•  (n-1)/(a_n^(2/3)+a_n^(1/3)+a_1^(2/3) )
•  (n-1)/(a_n^(2/3)+a_n^(1/3) a_1^(1/3)+a_1^(2/3) )
•  (n+1)/(a_(n+1)^(2/3)+a_(n+1)^(1/3) a_1^(1/3)+a_1^(2/3) )

Solution
Let d be the common difference of the A.P. Now, ∑_(r=1)^(n-1)▒1/((a_(r+1) )^(2/3)+(a_(r+1) )^(1/3) (a_r )^(1/3)+(a_r )^(2/3) ) =∑_(r=1)^(n-1)▒((a_(r+1) )^(1/3)-(a_r )^(1/3))/(a_(r+1)-a_r ) =1/d ∑_(r=1)^(n-1)▒〖{(a_(r+1) )^(1/3)-(a_r )^(1/3) } [∵a_(r+1)-a_r=d]〗 =1/d [(a_(n+1) )^(1/3)-(a_1 )^(1/3) ] =1/d×(a_n-a_1)/((a_n )^(2/3)+(a_n a_1 )^(1/3)+(a_1 )^(2/3) ) =(n-1)/(a_n^(2/3)+a_n^(1/3) a_1^(1/3)+a_1^(2/3) ) [∵a_n-a_1=(n-1) d]

Q9.If d,e,f are in G.P. and the two quadratic equations ax^2+2 bx+c=0 and dx^2+2 ex+f=0 have a common root, then

•  d/a,e/b,f/c are in H.P.
•  d/a,e/b,f/c are in G.P.
•  abf=aef+cde
•  b^2 df=ace^2

Solution
(a) We have, d,e,f are in G.P. ⇒e^2=df …(i) Now, dx^2+2ex+f=0 ⇒dx^2+2√df x+f=0 [Using (i)] ⇒(√d x+√f)^2=0⇒x=-√f/√d Putting x=-√f/√d in ax^2+2bx+c=0, we get a f/d+c=2b√(f/d) …(ii) ⇒a/b+c/f=(2 b)/√fd ⇒a/d+c/f=(2 b)/e [Using (i)] ⇒d/a,e/b,f/c are in H.P. Again from (ii), we have af+cd=2 b√fd ⇒aef+ced=2 be √fd⇒aef+cde=2 b df [Using (i)]

Q10. Consider the sequence of numbers 121, 12321, 1234321, … Each term in the sequence is

•  A prime number
•  Square of an odd number
•  Divisible by 11
•  Form a GP

Solution
The nth term of the sequence is …n(n+1)n…321 =1×10^2n+2×10^(2n-1)+3×10^(2n-2)+⋯+n×10^(n+1)+(n+1)×10^n ×10^(n-1)+...+3×10^2+2×10+1 Let S=1×10^2n+2×10^(2n-1)+...+n×10^n+1 (AG series) 1/10 S_1=1×10^(2n-1)+...+(n-1)×10^(n-1)+n×10^n On solving, we get 9/10 S_1=1×10^2n+...+1×10^(n+1)-n×10^n S_1=10/9 [10^2n+10^(2n-1)+⋯+10^(n+1)-n×10^n ] S_1=10/9 [(10^2n (1-1/10^n ))/(1-1/10)-n×10^n ] =(10/9)^2 [10^2n (1-1/10^n )]-[(10/9)n×10^n ] Again, let S_2=(n+1)×10^n+n×10^(n-1)+...+2×10+1 (1/10 S_2=(n+1)×10^(n-1)+...+3×10+2+1/10)/(9/10 S_2=(n+1)×10^n-[10^(n-1)+10^(n-2)+...+1+1/10] ) =(n+1)×10^n-(10^(n-1) (1-1/10^(n-1) ))/(1-1/10) ⇒ S_2=10/9 (n+1)×10^n-(10/9)^2 10^(n-1) (1-1/10^(n-1) ) ∴ The nth term = S_1+S_2 =(10/9)^2 [10^2n (1-1/10^(n-1) )-10^(n-1) (1-1/10^(n-1) )]+10/9 [-n×10^n+(n+1)×10^n ] =(10/9)^2 [10^2n-〖2.10〗^(n-1)+1/10^2 ] =10/9 [10^n-1/10]^2 =[(10^(n-1)-1)/9]^2 =[(999…(n+1)times)/9]^2 =[111…(n+1)times]^2=(odd number)^2