wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

List I has four entries and List II has five entries. Each entry of List I is to be matched with one or more than one entries of List II.

List IList II (A)If a,b,c,d are in A.P. such that a+b+c+d=32 and (P)a+b=815ad=7bc, then (a<b<c<d)(B)If the first three terms of the sequence 116,a,b,16 are in G.P.(Q)ab=12and the last three terms are in H.P., then (a>0,b>0)(C)If Sn=(3+1)2n+(31)2n,nN and (R)a+b=4Sn+1=aSn+bSn1, then(D)If aCb=84, aCb1=36 and aCb+1=126, then(S)a+b=12(T)a+d=16

Which of the following is the only CORRECT combination?

A
(C)(Q),(R)
loader
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
(C)(R),(S)
loader
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
(D)(P),(S)
loader
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
(D)(Q),(S)
loader
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A (C)(Q),(R)
(C)
Given, Sn=(3+1)2n+(31)2n
=(4+23)n+(423)n
=αn+βn,
where α=4+23, β=423
Now, Sn+1=αn+1+βn+1
=(α+β)(αn+βn)αβ(αn1+βn1)
Sn+1=8Sn4Sn1
a=8,b=4

(D)
Given aCb=84, aCb1=36, aCb+1=126
aCbaCb1=8436
ab+1b=73
3a10b+3=0 (1)

aCb+1aCb=12684
abb+1=32
2a5b3=0 (2)
From (1) and (2), we get
a=9,b=3

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Arithmetic Progression
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon
footer-image