12th Grade > Mathematics
SEQUENCES AND SERIES MCQs
Sequence And Series
Total Questions : 75
| Page 8 of 8 pages
Answer: Option B. -> 2
:
B
Suppose that the added number be x
then x + 2, x + 14, x + 62 be in G.P.
Therefore (x+14)2 = (x + 2)(x + 62)
⇒x2+196+28x = x2+64x+124
⇒36x = 72 ⇒x = 2
Trick : (a) Let 1 is added, then the numbers will be 3, 15, 63 which are obviously not in G.P.
(b) Let 2 is added, then the numbers will be 4, 16, 64 which are obviously in G.P.
:
B
Suppose that the added number be x
then x + 2, x + 14, x + 62 be in G.P.
Therefore (x+14)2 = (x + 2)(x + 62)
⇒x2+196+28x = x2+64x+124
⇒36x = 72 ⇒x = 2
Trick : (a) Let 1 is added, then the numbers will be 3, 15, 63 which are obviously not in G.P.
(b) Let 2 is added, then the numbers will be 4, 16, 64 which are obviously in G.P.
Answer: Option B. -> t+s−n
:
B
Given,
at=a+(t−1)d=s,...(i)
as=a+(s−1)d=t...(ii)
⇒(t−s)=a+(s−1)d−a−(t−1)d
⇒t−s=(s−t)d[from eqns. (i) and (ii)]
⇒(t−s)=−(t−s)d
⇒d=−1
From (i), we get
a+(t−1)(−1)=s.
⇒a=s+t−1
an=a+(n−1)d
⇒an=s+t−1+(n−1)×(−1)=s+t−1−n+1=s+t−n
:
B
Given,
at=a+(t−1)d=s,...(i)
as=a+(s−1)d=t...(ii)
⇒(t−s)=a+(s−1)d−a−(t−1)d
⇒t−s=(s−t)d[from eqns. (i) and (ii)]
⇒(t−s)=−(t−s)d
⇒d=−1
From (i), we get
a+(t−1)(−1)=s.
⇒a=s+t−1
an=a+(n−1)d
⇒an=s+t−1+(n−1)×(−1)=s+t−1−n+1=s+t−n
Answer: Option C. -> 0
:
C
Given,
8a8=15a15.
Since the nth term of an AP of first term a and common difference d is given by
an=a+(n−1)d, we have
8[a+(8−1)d]=15[a+(15−1)d].⇒8(a+7d)=15(a+14d)⇒8a+56d=15a+210d⇒7a+154d=0⇒a+22d=0⇒a+(23−1)d=0
Hence, the 23rd term of the AP is zero.
:
C
Given,
8a8=15a15.
Since the nth term of an AP of first term a and common difference d is given by
an=a+(n−1)d, we have
8[a+(8−1)d]=15[a+(15−1)d].⇒8(a+7d)=15(a+14d)⇒8a+56d=15a+210d⇒7a+154d=0⇒a+22d=0⇒a+(23−1)d=0
Hence, the 23rd term of the AP is zero.
Answer: Option B. -> 12
:
B
T5 = ar4 = 13 .......(i)
and T9 = ar8 = 16243 .......(ii)
Solving (i) and (ii), we get r = 23 and a = 2716
Now 4th term = ar3 = 3324.2333= 12
:
B
T5 = ar4 = 13 .......(i)
and T9 = ar8 = 16243 .......(ii)
Solving (i) and (ii), we get r = 23 and a = 2716
Now 4th term = ar3 = 3324.2333= 12
Answer: Option A. -> 1
:
A
a = ARp−1, b = ARq−1, c = ARr−1
∴(cb)p(ba)r(ac)q = (ARp−1ARq−1)p(ARq−1ARp−1)r(ARp−1ARr−1)q
= R(r−q)p+(q−p)r+(p−r)q = R0 = 1
:
A
a = ARp−1, b = ARq−1, c = ARr−1
∴(cb)p(ba)r(ac)q = (ARp−1ARq−1)p(ARq−1ARp−1)r(ARp−1ARr−1)q
= R(r−q)p+(q−p)r+(p−r)q = R0 = 1
Latest Videos
Latest Test Papers
Login With Google
Old way of Login
Login With Email