12th Grade > Mathematics
SEQUENCES AND SERIES MCQs
Sequence And Series
Total Questions : 75
| Page 4 of 8 pages
Answer: Option C. -> H.P
:
C
Since the roots of the quadratic equation in x are equal, we have (B2−4AC) = 0
⇒b2(c−a)2−4ac(b−c)(a−b) = 0
⇒b2(c2−2ca+a2)−4ac(ba−b2−ca+bc) = 0
⇒b2(c2+2ca+a2)−4ac{b(a+c)−ac}= 0
⇒b2(a+c)2−4ac{b(a+c)−ac}= 0
Which can be seen to be true, if
b = 2aca+c or b(a+c) = 2ac i.e., if a,b,c are in H.P.
:
C
Since the roots of the quadratic equation in x are equal, we have (B2−4AC) = 0
⇒b2(c−a)2−4ac(b−c)(a−b) = 0
⇒b2(c2−2ca+a2)−4ac(ba−b2−ca+bc) = 0
⇒b2(c2+2ca+a2)−4ac{b(a+c)−ac}= 0
⇒b2(a+c)2−4ac{b(a+c)−ac}= 0
Which can be seen to be true, if
b = 2aca+c or b(a+c) = 2ac i.e., if a,b,c are in H.P.
Answer: Option C. -> mnr
:
C
Given Tm = n, Tn = m for H.P. therefore for the corresponding A.P. mth term = 1n, nth term = 1m
Let a and d be the first term and common difference of this A.P., then
a+(m−1)d = 1n ......(i)
a+(n−1)d = 1m ......(i)
Solving these, we get a = 1mn, d = 1mn
Now, rth term of corresponding A.P.
= a+(r−1)d = 1mn+(r−1)1mn = 1+r−1mn = rmn
Therefore rth term of corresponding H.P. is mnr
Note : Students should remember this question as a fact.
:
C
Given Tm = n, Tn = m for H.P. therefore for the corresponding A.P. mth term = 1n, nth term = 1m
Let a and d be the first term and common difference of this A.P., then
a+(m−1)d = 1n ......(i)
a+(n−1)d = 1m ......(i)
Solving these, we get a = 1mn, d = 1mn
Now, rth term of corresponding A.P.
= a+(r−1)d = 1mn+(r−1)1mn = 1+r−1mn = rmn
Therefore rth term of corresponding H.P. is mnr
Note : Students should remember this question as a fact.
Answer: Option A. -> n(3n−1)2
:
A
π∑i=1 = 3π∑i=1i - 2π∑i=11 = 3
n(n+1)2 - 2n =
n(3n−1)2
:
A
π∑i=1 = 3π∑i=1i - 2π∑i=11 = 3
n(n+1)2 - 2n =
n(3n−1)2
Answer: Option B. -> A.P
:
B
If(b−c)2,(c−a)2,(a−b)2 are in A.P.
Then we have(c−a)2−(b−c)2 = (a−b)2−(c−a)2
⇒(b−a)(2c−a−b) = (c - b)(2a - b - c) ......(i)
Also if 1b−c,1c−a,1a−b are in A.P.
Then 1c−a−1b−c = 1a−b−1c−a
⇒b+a−2c(c−a)(b−c) = c+b−2a(a−b)(c−a)
⇒(a−b)(b+a−2c) = (b−c)(c+b−2a)
⇒(b−a)(2c−a−b) = (c−b)(2a−b−c)
Which is true by virtue of (i).
:
B
If(b−c)2,(c−a)2,(a−b)2 are in A.P.
Then we have(c−a)2−(b−c)2 = (a−b)2−(c−a)2
⇒(b−a)(2c−a−b) = (c - b)(2a - b - c) ......(i)
Also if 1b−c,1c−a,1a−b are in A.P.
Then 1c−a−1b−c = 1a−b−1c−a
⇒b+a−2c(c−a)(b−c) = c+b−2a(a−b)(c−a)
⇒(a−b)(b+a−2c) = (b−c)(c+b−2a)
⇒(b−a)(2c−a−b) = (c−b)(2a−b−c)
Which is true by virtue of (i).
Answer: Option A. -> G.P
:
A
If a and d be the first term and common difference of theA.P.
Then Tp = a + (p - 1)d, Tq = a + (q - 1)d and
Tr = a+(r - 1)d.
If Tp,Tq,Tr are in G.P.
Then its common ratio R = TqTp = TrTq = Tq−TrTp−Tq
= [a+(q−1)d]−[a+(r−1)d][a+(p−1)d]−[a+(q−1)d] = q−rp−q
Similarly, we can show that R =q−rp−q = r−sq−r
Hence (p - q), (q - r),(r - s) be in G.P.
:
A
If a and d be the first term and common difference of theA.P.
Then Tp = a + (p - 1)d, Tq = a + (q - 1)d and
Tr = a+(r - 1)d.
If Tp,Tq,Tr are in G.P.
Then its common ratio R = TqTp = TrTq = Tq−TrTp−Tq
= [a+(q−1)d]−[a+(r−1)d][a+(p−1)d]−[a+(q−1)d] = q−rp−q
Similarly, we can show that R =q−rp−q = r−sq−r
Hence (p - q), (q - r),(r - s) be in G.P.
Answer: Option C. -> G.P
:
C
Here logxloga,logxlogb,logxlogc are in H.P.
⇒logalogx,logblogx,logclogx are in A.P.
⇒logxa,logxb,logxc are in A.P.
⇒a,b,c are in G.P.
Note: Students should remember this question as afact
:
C
Here logxloga,logxlogb,logxlogc are in H.P.
⇒logalogx,logblogx,logclogx are in A.P.
⇒logxa,logxb,logxc are in A.P.
⇒a,b,c are in G.P.
Note: Students should remember this question as afact
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 D. -> 10
:
D
Series, 2,52,103,................. are in H.P. ⇒12,25,310,......... will be in A.P.
Now first term a = 12 and common difference d = −110
So, 5th term of the A.P. = 12+(5−1)(−110) = 110
Hence 5th term in H.P. is 10.
:
D
Series, 2,52,103,................. are in H.P. ⇒12,25,310,......... will be in A.P.
Now first term a = 12 and common difference d = −110
So, 5th term of the A.P. = 12+(5−1)(−110) = 110
Hence 5th term in H.P. is 10.
Answer: Option A. -> A.P
:
A
Let three number a,b and c in G.P., then b2 = ac
⇒2logdb = logda+logdc or logdb = logda+logdc2
Thus their logarithms are in A.P.
:
A
Let three number a,b and c in G.P., then b2 = ac
⇒2logdb = logda+logdc or logdb = logda+logdc2
Thus their logarithms are in A.P.
Answer: Option A. -> n(n+1)(n+2)3
:
A
Tπ = n2 + n ⇒Sπ = ∑Tπ = ∑ n2 +∑n
= n(n+1)(2n+1)6 + n(n+1)2
= n(n+1)6 {2n + 1 + 3} = n(n+1)(n+2)3
:
A
Tπ = n2 + n ⇒Sπ = ∑Tπ = ∑ n2 +∑n
= n(n+1)(2n+1)6 + n(n+1)2
= n(n+1)6 {2n + 1 + 3} = n(n+1)(n+2)3
Latest Videos
Latest Test Papers
Login With Google
Old way of Login
Login With Email