12th Grade > Mathematics
SEQUENCES AND SERIES MCQs
Sequence And Series
Total Questions : 75
| Page 5 of 8 pages
Answer: Option B. -> n(n+1)(3n2+7n+2)12
:
B
Tπ = n2(n + 1) = n3 + n2
Sπ = ∑Tπ =∑n3 + ∑n2 = [n(n+1)2]2+ n(n+1)(2n+1)6
= n(n+1)2[n(n+1)2 + 2n+13] = n(n+1)(3n2+7n+2)12
:
B
Tπ = n2(n + 1) = n3 + n2
Sπ = ∑Tπ =∑n3 + ∑n2 = [n(n+1)2]2+ n(n+1)(2n+1)6
= n(n+1)2[n(n+1)2 + 2n+13] = n(n+1)(3n2+7n+2)12
Answer: Option B. -> logba = logcb
:
B
x, y, z are in G.P., then y2 = x.z
Nowax =by = cz = m
⇒xlogda = ylogdb = zlogdc = logdm
⇒x = logam, y=logbm, z = logcm
Again as x, y, z are in G.P., so yx = zy
⇒logbmlogam = logcmlogbm ⇒logba = logcb
:
B
x, y, z are in G.P., then y2 = x.z
Nowax =by = cz = m
⇒xlogda = ylogdb = zlogdc = logdm
⇒x = logam, y=logbm, z = logcm
Again as x, y, z are in G.P., so yx = zy
⇒logbmlogam = logcmlogbm ⇒logba = logcb
Answer: Option B. -> 2√ab
:
B
We know that A > G > H
Where A is arithmetic mean, G is geometric mean and H is harmonic mean, then A > G
⇒a+b2>√ab or (a + b) > 2√ab
:
B
We know that A > G > H
Where A is arithmetic mean, G is geometric mean and H is harmonic mean, then A > G
⇒a+b2>√ab or (a + b) > 2√ab
Answer: Option A. -> | rp - 7| ≥ 4 √3
:
A
p,q,r are positive and are in A.P.
∴ q = p+r2 .........(i)
∵ The roots ofp x2 + qx + r = 0 are real
⇒q2≥ 4pr⇒ [p+r2]2≥ 4pr [using (i)]
⇒p2 +r2 - 14pr≥ 0
⇒ (rp)2 - 14 (rp) + 1≥ 0 ( ∵ p>0 and p≠ 0)
(rp−7)2 - (4√3)2≥ 0
⇒| rp - 7|≥ 4 √3.
:
A
p,q,r are positive and are in A.P.
∴ q = p+r2 .........(i)
∵ The roots ofp x2 + qx + r = 0 are real
⇒q2≥ 4pr⇒ [p+r2]2≥ 4pr [using (i)]
⇒p2 +r2 - 14pr≥ 0
⇒ (rp)2 - 14 (rp) + 1≥ 0 ( ∵ p>0 and p≠ 0)
(rp−7)2 - (4√3)2≥ 0
⇒| rp - 7|≥ 4 √3.
Answer: Option B. -> 7
:
B
Let four numbers are a−3d,a−d,a+d,a+3d.
Now (a−3d)+(a+3d)=8⇒a=4 and (a−d)(a+d)=15⇒a2−d2=15⇒d=1 Thus required numbers are 1,3,5,7. Hence greatest number is 7.
:
B
Let four numbers are a−3d,a−d,a+d,a+3d.
Now (a−3d)+(a+3d)=8⇒a=4 and (a−d)(a+d)=15⇒a2−d2=15⇒d=1 Thus required numbers are 1,3,5,7. Hence greatest number is 7.
Answer: Option D. -> None of these
:
D
we have 1a,1b,1c are in A.P. Let 1a = p - q, 1b = p and
1c = p+ q, where p,q > 0 and p > q. Now, substitute these
values in 3a+2b2a−b+3c+2b2c−b then it reduces to
10+14q2p2−q2 which is obviously greater than
10(as p > q > 0).
Trick: Put a = 1, b = 12, c = 13.
The expression has the value 3+12−12+1+123−12 = 83+12>0
:
D
we have 1a,1b,1c are in A.P. Let 1a = p - q, 1b = p and
1c = p+ q, where p,q > 0 and p > q. Now, substitute these
values in 3a+2b2a−b+3c+2b2c−b then it reduces to
10+14q2p2−q2 which is obviously greater than
10(as p > q > 0).
Trick: Put a = 1, b = 12, c = 13.
The expression has the value 3+12−12+1+123−12 = 83+12>0
Answer: Option A. -> 35
:
A
Given that a+ar+ar2a+ar+ar2+ar3+ar4+ar5=125152.
⇒1+r+r21+r+r2+r3+r4+r5=125152i.e.,1+r+r2(1+r+r2)+r3(1+r+r2)=125152⇒11+r3=125152⇒r3=152125−1=27125⇒r=35
:
A
Given that a+ar+ar2a+ar+ar2+ar3+ar4+ar5=125152.
⇒1+r+r21+r+r2+r3+r4+r5=125152i.e.,1+r+r2(1+r+r2)+r3(1+r+r2)=125152⇒11+r3=125152⇒r3=152125−1=27125⇒r=35
Answer: Option B. -> 129
:
B
Here first term of A.P. be 7 and second be 9, then 12th term will be 7 + 11 ×2 = 29
Hence 12th term of the H.P. be 129.
:
B
Here first term of A.P. be 7 and second be 9, then 12th term will be 7 + 11 ×2 = 29
Hence 12th term of the H.P. be 129.
Answer: Option A. -> √5−12
:
A
The given condition can be expressed as
Tn = Tn+1+Tn+2, where n≥1.
If a is the first term and r is the common ratio, then by the condition is
arn−1=arn+arn+1.⇒rn−1=rn(1+r)
⇒r2+r−1=0
⇒r=−1±√1+42=−1±√52
Since each term is positive, the common ratio of the GP is √5−12.
:
A
The given condition can be expressed as
Tn = Tn+1+Tn+2, where n≥1.
If a is the first term and r is the common ratio, then by the condition is
arn−1=arn+arn+1.⇒rn−1=rn(1+r)
⇒r2+r−1=0
⇒r=−1±√1+42=−1±√52
Since each term is positive, the common ratio of the GP is √5−12.
Answer: Option D. -> 149
:
D
Considering corresponding A.P.
a + 6d = 10 and a + 11d = 25 ⇒d= 3, a= -8
⇒T20 = a + 19d = -8 + 57 = 49
Hence 20th term of the corresponding H.P. is 149
:
D
Considering corresponding A.P.
a + 6d = 10 and a + 11d = 25 ⇒d= 3, a= -8
⇒T20 = a + 19d = -8 + 57 = 49
Hence 20th term of the corresponding H.P. is 149
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