12th Grade > Mathematics
SEQUENCES AND SERIES MCQs
Sequence And Series
Total Questions : 75
| Page 1 of 8 pages
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 A. -> 1−(n+1)xn+nxn+1(1−x)2
:
A
Let Sn be the sum of the given series to n terms, then
Sn = 1 + 2x + 3
x2 + 4
x3 + ....... + n
xn−1 ..........(i)
xSn = x + 2
x2 + 3
x2 + ...........+ n
xn ..........(ii)
Subtracting (ii) from (i), we get
(1-x)Sn = 1 + x +
x2 +
x3 + ......... to n terms -n
xn
= (
(1−xn)(1−x)) - nxn
⇒ Sn =
(1−xn)−nxn(1−x)(1−x)2 =
1−(n+1)xn+nxn+1(1−x)2
:
A
Let Sn be the sum of the given series to n terms, then
Sn = 1 + 2x + 3
x2 + 4
x3 + ....... + n
xn−1 ..........(i)
xSn = x + 2
x2 + 3
x2 + ...........+ n
xn ..........(ii)
Subtracting (ii) from (i), we get
(1-x)Sn = 1 + x +
x2 +
x3 + ......... to n terms -n
xn
= (
(1−xn)(1−x)) - nxn
⇒ Sn =
(1−xn)−nxn(1−x)(1−x)2 =
1−(n+1)xn+nxn+1(1−x)2
Answer: Option B. -> √mn
:
B
Given that
ap+q=arp+q−1=m and ap−q=arp−q−1=n.
⇒m×n=arp+q−1 x arp−q−1
=a2r2(p−1)
=(arp−1)2
⇒√mn=arp−1=ap
Thus, the pth term of the GP is√mn.
Aliter: Each term in a G.P. is the geometric mean of the terms equidistant from it. Here, (p+q)th and (p−q)th terms are equidistant from the pth term.
∴ pth term(√mn) will be G.M. of (p+q)th and (p−q)th terms.
:
B
Given that
ap+q=arp+q−1=m and ap−q=arp−q−1=n.
⇒m×n=arp+q−1 x arp−q−1
=a2r2(p−1)
=(arp−1)2
⇒√mn=arp−1=ap
Thus, the pth term of the GP is√mn.
Aliter: Each term in a G.P. is the geometric mean of the terms equidistant from it. Here, (p+q)th and (p−q)th terms are equidistant from the pth term.
∴ pth term(√mn) will be G.M. of (p+q)th and (p−q)th terms.
Answer: Option B. -> nn+1
:
B
(
11 -
12) + (
12 -
13) + (
13 -
14) + .........+ (
1n -
1n+1)
= 1 -
1n+1 =
nn+1
:
B
(
11 -
12) + (
12 -
13) + (
13 -
14) + .........+ (
1n -
1n+1)
= 1 -
1n+1 =
nn+1
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
Answer: Option A. -> x,y,z will be in A.P.
:
A
Leta1x=b1y=c1z=k.
⇒a=kx,b=ky,c=kz.
Sincea,b,c are in G.P., b2=ac.
∴k2y=kx.kz=kx+z
∴2y=x+z
⇒x,y,z are in A.P.
:
A
Leta1x=b1y=c1z=k.
⇒a=kx,b=ky,c=kz.
Sincea,b,c are in G.P., b2=ac.
∴k2y=kx.kz=kx+z
∴2y=x+z
⇒x,y,z are in A.P.
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 A. -> 3ad
:
A
Since a, b, c, d are in H.P., therefore b is the H.M. of a and c
i.e., b = 2acb+d, ∴(a+c)(b+d) = 2acb.2bdc
⇒ab+ad+bc+cd = 4ad ⇒ab+bc+cd = 3ad
Trick : Check for a =1, b = 12, c = 13, d = 14
:
A
Since a, b, c, d are in H.P., therefore b is the H.M. of a and c
i.e., b = 2acb+d, ∴(a+c)(b+d) = 2acb.2bdc
⇒ab+ad+bc+cd = 4ad ⇒ab+bc+cd = 3ad
Trick : Check for a =1, b = 12, c = 13, d = 14
Answer: Option D. -> All of these
:
D
Let S = 3 + 7 + 13 + 21 + ...... + Tπ ⇒Tπ = n2 + n + 1.
LetTr = cot−1( r2 + r + 1) = tan−1(r + 1) - tan−1r.
Put r = 1, 2,........, n and add, we get the required sum
tan−1(n+1) - tan−11 = tan−1( nn+2) =cot−1 (n+2n)
:
D
Let S = 3 + 7 + 13 + 21 + ...... + Tπ ⇒Tπ = n2 + n + 1.
LetTr = cot−1( r2 + r + 1) = tan−1(r + 1) - tan−1r.
Put r = 1, 2,........, n and add, we get the required sum
tan−1(n+1) - tan−11 = tan−1( nn+2) =cot−1 (n+2n)
Answer: Option C. -> 16
:
C
T6 of H.P. = 161⇒T6 of A.P. = 61
and T10 of H.P. = 1105⇒T10 of A.P. = 105,
so, a + 5d = 61 ........(i) and a + 9d = 105 .....(ii)
From (i) and (ii), a= 6
Therefore first term of H.P. = 1a = 16.
:
C
T6 of H.P. = 161⇒T6 of A.P. = 61
and T10 of H.P. = 1105⇒T10 of A.P. = 105,
so, a + 5d = 61 ........(i) and a + 9d = 105 .....(ii)
From (i) and (ii), a= 6
Therefore first term of H.P. = 1a = 16.
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