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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 log_2⁡〖a+log_4⁡〖b+log_4⁡〖c=2〗 〗 〗 log_9⁡〖a+log_3⁡〖b+log_9⁡〖c=2〗 〗 〗 log_16⁡〖a+log_16⁡〖b+log_4⁡〖c=2,〗 〗 〗 then
•  a=2/3,b=27/8,c=32/3
•  a=27/8,b=2/3,c=32/3
•  a=32/3,b=32/3,c=2/3
•  a=2/3,b=32/3,c=27/3
Solution
(a) We have, log_2⁡〖a+log_4⁡〖b+log_4⁡〖c=2〗 〗 〗 log_9⁡〖a+log_3⁡〖b+log_9⁡〖c=2〗 〗 〗 log_16⁡〖a+log_16⁡〖b+log_4⁡〖c=2〗 〗 〗 ⇒log_2⁡〖a+1/2 log_2⁡〖b+1/2 log_2⁡〖x=2〗 〗 〗 ⇒1/2 log_3⁡〖a+log_3⁡〖b+1/2 log_3⁡〖c=2〗 〗 〗 ⇒1/2 log_4⁡〖a+1/2 log_4⁡〖b+log_4⁡〖c=2〗 〗 〗 ⇒log_2⁡〖(a^2 bc)=4,〗 log_3⁡〖(ab^2 c)〗=4,log_4⁡〖(abc^2)〗=4 ⇒a^2 bc=2^4,ab^2 c=3^4 and abc^2=4^4 ⇒(a^2 bc)(ab^2 c)(abc^2 )=2^4×3^4×4^4 ⇒(abc)^4=(2×3×4)^4⇒abc=24 Now, a^2 bc=2^4 and abc=24⇒a=16/24=2/3 ab^2 c=3^4 and abc=24⇒b=81/21=27/8 abc^2=4^4 and abc=24⇒c=256/24=32/3

Q2.If an infinite geometric series the first term is a and common ratio is r. If the sum of the series is 4 and the second term is 3/4, then (a,r) is
•  (4/7,3/7)
•  (2,3/8 )
•  (3/2,1/2)
•  (3,1/4)
Solution
Since, a/(1-r)=4 ⇒a=4(1-r) …(i) and ar=3/4 ⇒4(1-r)r=3/4 [from Eq. (i)] ⇒16r^2-16r+3=0 ⇒(4r-1)(4r-3)=0 ⇒ r=1/4,3/4 If r=1/4, then a=3

Q3.  If the sum of first n natural numbers is 1/5 times the sum of their squares, then the value of n is
•   5
•  6
•  7
•  8
Solution
Since, Σn=(1/5)Σn^2 ⇒(n(n+1))/2=1/5 (n(n+1)(2n+1))/6 ⇒2n+1=15⇒n=7

Q4. An infinite GP has first term x and sum 5, then x belongs to
•  x<-10 span="">
•  -10
•   0
•   x>10
Solution
Let r be the common ratio of give GP. ∴ x/(1-r)=5 ⇒ r=(1-x/5) ∵ For infinite GP, |r|< 1 ⇒ -1<1-x 10="">x>0 ⇒ 0

Q5.The sum of first 10 terms of the series (x+1/x)^2+(x^2+1/x^2 )^2+(x^3+1/x^3 )^2+⋯ is
•  ((x^20-1)/(x^2-1))((x^22+1)/x^20 )+20
•  ((x^18-1)/(x^2-1))((x^11+1)/x^9 )+20
•  ((x^18-1)/(x^2-1))((x^11-1)/x^9 )+20
•  None of these
Solution
We have, (x+1/x)^2+(x^2+1/x^2 )^2+(x^3+1/x^3 )^3+⋯+(x^10+1/x^10 )^2 =(x^2+x^4+x^6+⋯+x^20 )+(1/x^2 +1/x^4 +1/x^6 +⋯+1/x^20 )+20 x^2 ((x^20-1))/((x^2-1))+1/x^2 ((1-1/x^20 ))/((1-1/x^2 ) )+20 =((x^20-1)/(x^2-1))((x^22+1)/x^20 )+20

Q6. The values of 3 log⁡〖81/80+5 log⁡〖25/24+7 log⁡〖16/15〗 〗 〗 is
•  log 2
•  log 3
•  1
•  0
Solution
We have, 3 log⁡〖81/80+5 log⁡〖25/24+7 log⁡〖16/15〗 〗 〗 =3 log⁡(3^4/(2^4×5))+5 log⁡〖(5^2/(2^3×3))+7 log⁡(2^4/(3×5)) 〗 =log⁡{(3^4/(2^4×5))^3×(5^2/(2^3×3))^5×(2^4/(3×5))^7 } =log⁡〖{3^12/(2^12×5^3 )×5^10/(2^15×3^5 )×28/(3^7×5^7 )}=log⁡2 〗

Q7.If x_1,x_2,x_3 as well as y_1,y_2,y_3 are in GP with the same common ratio, then the points (x_1,y_1 ),(x_2,y_2) and (x_3,y_3)
•  Lie on a staright line
•  Lie on an ellipse
•  Lie on a circle
•  Are vertices of a triangle
Solution
Since, x_1,x_2,x_3 and y_1,y_2,y_3 are in GP with the same common ratio. ∴ x_2=rx_1,x_3=r^2 x_1,y_2=ry_1,y_3=r^2 y_1 Area of triangle =1/2 |■(x_1&y_1&1@x_2&y_2&1@x_3&y_3&1)|=1/2 |■(x_1&y_1&1@rx_1&ry_1&1@r^2 x_1&r^2 y_1&1)| =1/2 x_1 y_1 |■(1&1&1@r&r&1@r^2&r^2&1)|=0 (∵ two columns are identical) Hence, three points are in a straight line.

Q8.Three numbers whose sum is 15 are in AP. If they are added by 1, 4 and 19 respectively they are in GP. The numbers are
•  2,5,8
•  26,5,-16
•  2,5,8and26,5,-16
•  None of these
Solution
Let the three numbers in AP are a-d,a,a+d Since, a-d+a+a+d=15 ⇒a=5 Since, (a-d+1),(a+4),(a+d+19) are in GP. ∴(a+4)^2=(a-d+1)(a+d+19) ⇒ 9^2=(6-d)(24+d) [∵a=5] ⇒ d^2+18d-63=0 ⇒ d=3,-21 ∴ Required series are 2, 5, 8 and 26, 5, -16

Q9.The coefficients of x^6 in the expansion of log⁡〖{(1+x)^(1+x) (1-x)^(1-x)}〗, is
•  1/15
•  1/30
•  1/10
•  1/45
Solution
We have, Coefficient of x^6=2(1/5.6)=1/15

Q10. If a,b,c be in GP, then log⁡〖a^n,log⁡〖b^n,log⁡〖c^n 〗 〗 〗 will be
•  AP
•  GP
•  HP
•  None of these
Solution
If a,b,c are in GP, then b^2=ac Taking log on both sides, we get 2 log_e⁡〖b=log_e⁡〖a+log_e⁡c 〗 〗 ⇒2n log_e⁡〖b=n log_e⁡〖a+n log_e⁡c 〗 〗 ⇒2 log_e⁡〖b^n=log_e⁡〖a^n+log_e⁡〖c^n 〗 〗 〗 ⇒log_e⁡〖a^n,log_e⁡〖b^n,〗 〗 log_e⁡〖c^n 〗 be in AP. #### Written by: AUTHORNAME

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