SEQUENCE AND SERIES QUIZ -6

SEQUENCE AND SERIES QUIZ-6

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. If (x+y)/(1-xy),y,(y+z)/(1-yz) be in A.P., then x,1/y,z will be in

•  A.P
•  G.P
•  H.P
•  None of these

Solution
It is given that (x+y)/(1-xy),y,(y+z)/(1-yz) are in A.P. ⇒y-(x+y)/(1-xy)=(y+z)/(1-yz)-y ⇒(y-xy^2-x-y)/(1-xy)=(y+z-y+y^2 z)/(1-yz) ⇒-x/(1-xy)=z/(1-yz) ⇒-x+xyz=z-xyz ⇒2 xyz=x+z ⇒y=(x+z)/(2 xz) ⇒1/y=(2 xz)/(x+z) ⇒x,1/y,z are in H.P.

Q2.The interior angles of a polygon are in AP. If the smallest angle be 120° and the common difference be 5, then the number of side is

•  8
•  10
•  9
•  6

Solution
Let the number of sides of the polygon be n.Then, the sum of interior angles of the polygon =(2n-4) π/4=(n-2)π Since, the angles are in AP and a=120°,d=5 Therefore, S_n=n/2[2a+(n-1)d] ⇒n/2 [2×120+(n-1)5]=(n-2)180 ⇒n^2-25n+144=0 ⇒(n-9)(n-16)=0 ⇒n=9,16 Take n=16 T_16=a+15d=120°+15(5°)=195°, which is impossible, an interior angle cannot be greater than 180°. Hence, n=9

Q3.  If S=1/(1∙2)-1/(2∙3)+1/(3∙4)-1/(4∙5)+⋯+∞, then e^S equals

•   log_e⁡(4/e)
•  4/e
•  log_e⁡(e/4)
•  e/4

Solution
We have, S=1/(1∙2)-1/(2∙3)+1/(3∙4)-1/(4∙5)+⋯∞ ⇒S=(1/1-1/2)-(1/2-1/3)+(1/3-1/4)-(1/4-1/5)+⋯∞ ⇒S=2(1-1/2+1/3-1/4+⋯)-1 ⇒S=2 log⁡〖(1+1)-log_e⁡〖e=log_e⁡(4/e) 〗 〗 ∴e^S=4/e

Q4. If log_8⁡{log_2⁡(log_3⁡(x^2-4x+85) ) }=1/3, then x equals to

•  5
•  4
•  3
•  2

Solution
We have, log_8⁡〖[log_2⁡{log_3⁡(x^2-4x+85) } ]=1/3〗 ⇒log_2⁡〖{log_3⁡(x^2-4x+85) }=8^(1/3) 〗=2 ⇒log_3⁡〖(x^2-4x+85)=2^2 〗 ⇒x^2-4x+85=3^4 ⇒x^2-4x+4=0⇒(x-2)^2=0⇒x=2

Q5.If 1/1^2 +1/2^2 +1/3^2 +⋯to ∞=π^2/6, then 1/1^2 +1/3^2 +1/5^2 +⋯ equals

•  π^2/8
•  π^2/12
•  π^2/3
•  π^2/2

Solution
 We have, 1/1^2 +1/3^2 +1/5^2 +1/7^2 +⋯ =(1/1^2 +1/2^2 +1/3^2 +1/4^2 +1/5^2 +1/6^2 +1/7^2 +⋯)-(1/2^2 +1/4^2 +1/6^2 +⋯) π^2/6-1/4 (1/1^2 +1/2^2 +1/3^2 +⋯)=π^2/6-1/4 (π^2/6)=π^2/8

Q6. The product of n positive numbers is unity. Their sum is

•  A positive integer
•  Equal to n+(1/n)
•  Divisible by n
•  Never less than n

Solution
Given that, x_1 x_2 x_3…x_n=1 …(i) We know that, AM ≥ GM ∴ ((x_1+x_2+x_3+...+x_n)/n)≥(x_1 x_2 x_3…x_n )^(1/n) =(1)^(1/n)=1 [from Eq.(i)] ⇒ x_1+x_2+x_3+...+x_n≥n ∴ x_1+x_2+x_3+...+x_n Can never be less than n

Q7.Consider the following statement : 1. If mth term of HP is n and the nthe term is m, then the (mn)th term is 1. 2. If a,b,c are in AP and a,2b,c are in GP, then a,4b,c are in HP. 3. If any odd number of quantities are in AP, then first middle and last are in AP Which of the statement give above is/are correct?

•  Only(1)
•  Only(2)
•  Only(3)
•  All of these

Solution

Q8.If x,y,z are in HP, then log⁡〖(x+z)〗+log⁡〖(x-2y+z)〗 is equal to

•  log(x-z)
•  2log(x-z)
•  3log(x-z)
•  4log(x-z)

Solution
Since, y=2xz/(x+z) Now, x-2y+z=x+z-2(2xz/(x+z)) =x+z-4xz/(x+z)=(x-z)^2/(x+z) ⇒log⁡〖(x-2y+z)〗=log⁡〖(x-z)^2 〗-log⁡〖(x+z)〗 ⇒log⁡〖(x-2y+z)〗+log⁡〖(x+z)〗=2 log⁡〖(x-z)〗

Q9.Number of values of x for which [x], sgn x,{x}{x≠0} are in AP, is

•  0
•  2
•  3
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Solution

Q10. In a geometric progression (GP) the ratio of the sum of the first three terms and first six terms is 125:152 the common ratio is

•  1/5
•  2/5
•  4/5
•  3/5

Solution
Since, (a+ar+ar^2)/(a+ar+ar^2+ar^3+ar^4+ar^5 )=125/162 ⇒(1+r+r^2)/(1+r+r^2 )(1+r^3 ) =125/162 ⇒1+r^3=152/125 ⇒r^3=27/125=(3/5)^3 ⇒ r=3/5