SEQUENCE AND SERIES QUIZ-3

SEQUENCE AND SERIES QUIZ-3

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 a,b,c are in G.P., then log_a⁡x,log_b⁡〖x,log_c⁡x 〗 are in

•  A.P
•  G.P
•  H.P
•  None of these

Solution
Since a,b,c are in G.P. ∴b^2=ac ⇒2 log_x⁡〖b=log_x⁡〖a+log_x⁡c 〗 〗 ⇒2/log_b⁡x =1/log_a⁡x +1/log_c⁡x ⇒log_a⁡〖x,〗 log_b⁡〖x,〗 log_c⁡x are in H.P.

Q2.The difference between two numbers is 48 and the difference between their arithmetic mean and their geometric mean is 18. Then, the greater of two numbers is

•  96
•  60
•  54
•  49

Solution
Let the two numbers be a and b ∴ a-b=48 and (a+b)/2-√ab=18 ⇒(√a-√b)(√a+√b)=48 and (√a-√b)=6 ⇒√a+√b=8 and (√a-√b)=6 ⇒√a=7 and √b=1 ⇒a=49 and b=1 Hence, the greater number is 49.

Q3.   If log⁡〖(2a-3b)〗=log⁡〖a-log⁡〖b,〗 〗 then a=

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

Solution
We have, log⁡〖(2a-3b)=log⁡〖a-log⁡b 〗 〗 ⇒log⁡〖(2a-3b)〗=log⁡(a/b) ⇒2a-3b=a/b⇒2ab-3b^2=a⇒a=(3b^2)/(2b-1)

Q4. The sum of the infinite series 1+1/2!+1.3/4!+1.3.5/6!+⋯ is

•  e
•  e^2
•  √e
•  1/e

Solution
Let S=1+1/2!+1.3/4!+1.3.5/6!+...∞ ∴T_n=(1.3.5.….(2n-1))/(2n)!×(2.4.….2n)/(2.4.….2n) =(2n)!/((2n)!2^n (n)!)=1/(2^n (n)!) ∴S=1+∑▒〖T_n=1+1/2(1)!+1/(2^2 (2)!)+...∞〗 =e^(1/2)=√e

Q5.The value of 1/2!+2/3!+...+999/1000! is equal to

•  (1000!-1)/1000!
•  (1000!+1)/1000!
•  (999!-1)/999!
•  (999!+1)/999!

Solution
 1/2!+2/3!+...+999/1000! =(2-1)/2!+(3-1)/3!+...+(1000-1)/1000! =1/1!-1/2!+1/2!-1/3!+...+1/999!-1/1000! 1-1/1000!=(1000!-1)/1000!

Q6. α,β are the roots of the equation x^2-3x+a=0 and γ,δ are the roots of the equation x^2-12x+b=0.If α,β,γ,δ form an increasing GP, then (a,b) is equal to

•  (3,12)
•  (12,3)
•  (2,32)
•  (4,16)

Solution

Q7.If the 7th term of HP is 1/10 and the 12th term is 1/25, then the 20th term is

•  1/41
•  1/45
•  1/49
•  1/37

Solution
The corresponding terms of HP in terms of AP is 10 and 25 ∴T_7=a+6d=10 … (i) and T_12=a+11d=25 …(ii) On solving Eqs. (i) and (ii), we get a=-8,d=3 ∴ T_20=-8+(20-1)3=49 Then, 20th term of HP is 1/49.

Q8.The value of 1-log⁡2+(log⁡2 )^2/2!-(log⁡2 )^3/3!+⋯ is

•  log⁡3
•  log⁡2
•  1/2
•  None of these

Solution
∴1-log⁡〖2+(〖log⁡2〗^ )^2/2!-(log⁡2 )^3/3!+⋯=e^(-log⁡2 )⁡ 〗 =e^log⁡〖2^(-1) 〗 =1/2

Q9.The HM of two numbers is 4. Their AM is A and GM is G. If 2A+G^2=27, then A is equal to

•  9
•  9/2
•  18
•  27

Solution
(b) Let the two numbers be a and b Since, 2ab/(a+b)=4, A=(a+b)/2 and G=√ab ∴ 2ab/2A=4 ⇒A=ab/4 and G^2=4A Given, 2A+G^2=27 ∴ 2A+4A=27 ⇒ A=9/2

Q10. If |a|<1 a="" and="" b="" is="" of="" series="" span="" sum="" the="" then="" upto="">

•  a/(1-a)+ab/(1-ab)
•  a^2/(1-a^2 )+ab/(1-ab)
•  b/(1-b)+a/(1-a)
•  b^2/(1-b^2 )+ab/(1-ab)

Solution
We have, |a|<1 a="" ab="" b="" now="" span="" to="" upto="">