SEQUENCE AND SERIES QUIZ-8

SEQUENCE AND SERIES QUIZ-8

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,1,z are in A.P. and x,2,z are in G.P., then x,4,z are in

•  A.P
•  G.P
•  H.P
•  None of the these

Solution
We have, ├ █(x,1,z are in AP⇒2=x+z@x,2,z are in GP⇒4=xz)} …(i) Since (i) does not satisfy 8=x+z and 16=xz. But, it satisfies the relation 4=(2 xz)/(x+2). Hence, x,4,z are in HP

Q2.The correct statement is

•  0.5+0.55+0.555+... to n terms =5n/9-5/81(1-10^(-n))
•  8+88+888+...+ to n terms=80/81 (10^n-1)-8n/9
•  1^2+(1^2+2^2 )+(1^2+2^2+3^2 )+...to n terms =(n(n+1)^2 (n+2))/12
•   All are correct

Solution
(d) We have, (a) 0.5+0.55+0.555+...=5/9 [0.9+0.99+0.999+...] =5/9[(1-0.1)+(1-0.01)+(1-0.001)+...to n terms] =5/9[(1+1+...to n terms)-├ (1/10+1/10^2 +1/10^3 +...+to n term)] =5/9 [n-(1/10 {1-1/10^n })/(1-1/10)] =5n/9-5/81 (1-10^(-n) ) (b) 8+88+888+...to n terms =8/9[9+99+999+...to n terms] =8/9[(10-1)+(10^2-1)+(10^3-1)+...to n terms] =8/9[(10+10^2+10^3+⋯+to n terms)-(1+1+1+…+to n terms)] =8/9 [10(10^n-1)/(10-1)-n] =80/81 (10^n-1)-8n/9 (c) the nth terms in the sequence is x_n=1^2+2^2+3^2+...+n^2 =(n(n+1)(2n+1))/6 =1/3 n^3+1/2 n^2+1/6 n ∴ The required sum =∑▒x_n =1/3 ∑▒〖n^3+1/2〗 ∑▒〖n^2+1/6〗 ∑▒n =1/3 [n(n+1)/2]^2+1/2.(n(n+1)(2n+1))/6+1/6.n(n+1)/2 =n(n+1)/12[(n+1)+(2n+1)+1 =n(n+1)/2 [n^2+3n+2] =n(n+1)/2(n+1)(n+2) =(n(n+1)^2 (n+2))/12

Q3.  The value of log_a⁡〖(log_b⁡x)〗/log_b⁡(log_a⁡b ) is

•   log_b⁡a
•  log_a⁡b
•   -log_a⁡b
•   -log_b⁡a

Solution
(c) We have, log_a⁡〖(log_b⁡a)〗/log_b⁡〖(log_a⁡b)〗 =log⁡〖(log_b⁡a)〗/log⁡a ×log⁡b/log⁡〖(log_a⁡b)〗 =log⁡(log⁡a/log⁡b )/log⁡a ×log⁡b/log⁡(log⁡b/log⁡a ) =log⁡〖(log⁡a )-log⁡〖(log⁡b)〗 〗/log⁡a ×log⁡b/log⁡〖(log⁡b )-log⁡〖(log⁡a)〗 〗 =-log⁡b/log⁡a =-log_a⁡b

Q4. If log_12⁡〖27=a,〗 then log_6⁡〖16=〗

•  (3-a)/(3+a)
•  4((3-a)/(3+a))
•  3((4-a)/(4+a))
•  3((4+a)/(4-a))

Solution
(b) We have, log_12⁡〖27=a〗 ⇒log_12⁡〖3^3=a〗 ⇒3 log_12⁡〖3=a〗 ⇒3/a=log_3⁡12 ⇒3/a=log_3⁡〖(2^2×3)=2 log_3⁡〖2+log_3⁡3 〗 〗 ⇒3/a=2 log_3⁡〖2+1〗 ⇒(3-a)/a=2 log_3⁡〖2⇒log_2⁡〖3=2a/(3-a)〗 〗 …(i) Now, log_6⁡〖16=log_6⁡〖2^4=4 log_6⁡〖2=4/log_2⁡6 〗 〗 〗 ⇒log_6⁡〖16=4/log_2⁡〖3+log_2⁡2 〗 〗=4/(2a/(3-a)+1) [Using (i)] ⇒log_6⁡〖16=4((3-a)/(3+a))〗

Q5.Let the sequence, a_1,a_2,a_3,…,a_2n, form an AP, then a_1^2-a_2^2+a_3^2-...+a_(2n-1)^2-a_2n^2 is equal to

•  n/(2n-1)(a_1^2-a_2n^2)
•  2n/(n-1)(a_2n^2-a_1^2)
•  n/(n+1)(a_1^2+a_2n^2)
•  None of these

Solution
 Since, a_1,a_2,a_3,…,a_n form an AP. ∴ a_2-a_1=a_4-a_3=...=a_2n-a_(2n-1)=d Let S=a_1^2-a_2^2+a_3^2-a_4^2+...+a_(2n-1)^2-a_2n^2 =(a_1-a_2 )(a_1+a_2 )+(a_3-a_4 )(a_3+a_4 )+...+(a_(2n-1)-a_2n )(a_(2n-1)+a_2n) = -d(a_1+a_2+...+a_2n )=-d(2n/2 (a_1+a_2n )) …(i) Also, we know a_2n=a_1+(2n-1)d ⇒ d=(a_2n-a)/(2n-1) ⇒-d=(a_1-a_2n)/(2n-1) On putting the value of d in Eq, (i), we get S=(n(a_1-a_2n )(a_1+a_2n))/(2n-1)=n/(2n-1)(a_1^2-a_2n^2)

Q6. Which of the following statement is correct?

•  If each earn of an AP a number is added or subtracted, then the series so obtained is also an AP.
•  The nth term of geometric series whose first term is a and common ratio r,is ar^(n-1).
•  If each term of a GP be raised to the same power the resulting terms are in GP.
•  All of the above

Solution

Q7.If log⁡〖2,〗 log⁡〖(2^x-1)〗 and log⁡〖(2^x+3)〗 are in A.P., then 〖2,2〗^x-〖1,2〗^x+3 are in

•  A.P
•  H.P
•  G.P
•  None of these

Solution
We have, log⁡〖2,〗 log⁡〖(2^x-1),〗 log⁡〖(2^x+3)〗 are in A.P. ⇒〖2,2〗^x-〖1,2〗^x+3 are in G.P.

Q8.If x^18=y^21=z^28, then 3,3 log_y⁡〖x,3 log_z⁡〖y,7 log_x⁡z 〗 〗 are in

•  A.P
•  G.P
•  H.P
•  None of these

Solution
Let x^18=y^21=z^28=k Then, 18 log⁡〖x=21 log⁡〖y=28 log⁡〖z=log⁡k 〗 〗 〗 ⇒log_y⁡〖x=21/18,log_z⁡〖y=28/21,log_x⁡〖z=18/28〗 〗 〗 ⇒3 log_y⁡〖x=7/2,3 log_z⁡〖y=4,7 log_x⁡z=9/2〗 〗 ⇒3,3 log_y⁡〖z,3 log_z⁡〖y,7 log_x⁡z 〗 〗 are in A.P.

Q9.The sum of n terms of an A.P. is a n(n-1). The sum of the squares of these terms is

•  a^2 n^2 (n-1)^2
•  a^2/6 n(n-1)(2n-1)
•   (2a^2)/3 n(n-1)(2n-1)
•  (2a^2)/3 n(n+1)(2n+1)

Solution
Let A be the first term and D be the common difference of the AP. Then, S_n=an(n-1) ⇒n/2 {2 A+(n-1)D}=an (n-1) ⇒2A+(n-1)D=2 an-2 a ⇒2A-D=-2 a and D=2 a ⇒A=0,D=2 a The sum of the squares of the n terms of the sequence is S=A^2+(A+D)^2+(A+2D)^2+⋯+{A+(n-1)D}^2 ⇒S=D^2 {1^2+2^2+3^2+⋯+(n-1)^2} ⇒S=4 a^2 (n(n-1)(2n-1))/6=(2 a^2)/3 n(n-1)(2n-1)

Q10. A G.P. consists of an even number of terms. If terms sum of all the terms is 5 times the sum of the terms occupying odd places, the common ration will be equal to

•  2
•  3
•  4
•  5

Solution
Let there be 2n terms in the given G.P. with first term a and the common ratio r. Then, Sum of all terms =5 (Sum of odd terms) ⇒a_1+a_2+⋯+a_2n=5(a_1+a_3+⋯+a_(2n-1) ) ⇒a+ar+ar^2+⋯+ar^(2n-1)=5(a+ar^2+⋯+ar^(2n-2) ) ⇒a ((r^(2 n)-1))/((r-1))=5a ((r^(2 n)-1))/((r^2-1)) ⇒r+1=5 ⇒r=4