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Class 11 Maths
Sequences And Series
Quiz 1
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If a, b, c are in G.P., then the equations ax² + 2bx + c = 0 and dx² + 2ex + f = 0 have a common root if d/a, e/b, f/c are in
0%
AP
0%
GP
0%
HP
0%
none of these
Explanation
AP
Hint:
Given a, b, c are in GP
⇒ b² = ac
⇒ b² – ac = 0
So, ax² + 2bx + c = 0 have equal roots.
Now D = 4b² – 4ac
and the root is -2b/2a = -b/a
So -b/a is the common root.
Now,
dx² + 2ex + f = 0
⇒ d(-b/a)² + 2e×(-b/a) + f = 0
⇒ db2 /a² – 2be/a + f = 0
⇒ d×ac /a² – 2be/a + f = 0
⇒ dc/a – 2be/a + f = 0
⇒ d/a – 2be/ac + f/c = 0
⇒ d/a + f/c = 2be/ac
⇒ d/a + f/c = 2be/b²
⇒ d/a + f/c = 2e/b
⇒ d/a, e/b, f/c are in AP
If a, b, c are in AP then
0%
b = a + c
0%
2b = a + c
0%
b² = a + c
0%
2b² = a + c
Explanation
2b = a + c
Hint:
Given, a, b, c are in AP
⇒ b – a = c – b
⇒ b + b = a + c
⇒ 2b = a + c
Three numbers form an increasing GP. If the middle term is doubled, then the new numbers are in Ap. The common ratio of GP is
0%
2 + √3
0%
2 – √3
0%
2 ± √3
0%
None of these
Explanation
2 + √3
Hint:
Let the three numbers be a/r, a, ar
Since the numbers form an increasing GP, So r > 1
Now, it is given that a/r, 2a, ar are in AP
⇒ 4a = a/r + ar
⇒ r² – 4r + 1 = 0
⇒ r = 2 ± √3
⇒ r = 2 + √3 {Since r > 1}
The sum of n terms of the series (1/1.2) + (1/2.3) + (1/3.4) + …… is
0%
n/(n+1)
0%
1/(n+1)
0%
1/n
0%
None of these
Explanation
n/(n+1)
Hint:
Given series is:
S = (1/1·2) + (1/2·3) + (1/3·4) – ………………. 1/n.(n+1)
⇒ S = (1 – 1/2) + (1/2 – 1/3) + (1/3 – 1.4) -……… (1/n – 1/(n+1))
⇒ S = 1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 – ……….. 1/n – 1/(n+1)
⇒ S = 1 – 1/(n+1)
⇒ S = (n + 1 – 1)/(n+1)
⇒ S = n/(n+1)
If 1/(b + c), 1/(c + a), 1/(a + b) are in AP then
0%
a, b, c are in AP
0%
a², b², c² are in AP
0%
1/1, 1/b, 1/c are in AP
0%
None of these
Explanation
a², b², c² are in AP
Hint:
Given, 1/(b + c), 1/(c + a), 1/(a + b)
⇒ 2/(c + a) = 1/(b + c) + 1/(a + b)
⇒ 2b² = a² + c²
⇒ a², b², c² are in AP
The sum of series 1/2! + 1/4! + 1/6! + ….. is
0%
e² - 1 / 2
0%
(e - 1)² /2 e
0%
e² - 1 / 2 e
0%
e² - 2 / e
Explanation
(e - 1)² /2 e
Hint:
We know that,
e
x
= 1 + x/1! + x² /2! + x³ /3! + x
4
/4! + ...........
Now,
e
1
= 1 + 1/1! + 1/2! + 1/3! + 1/4! + ...........
e
-1
= 1 - 1/1! + 1/2! - 1/3! + 1/4! + ...........
e
1
+ e
-1
= 2(1 + 1/2! + 1/4! + ...........)
⇒ e + 1/e = 2(1 + 1/2! + 1/4! + ...........)
⇒ (e² + 1)/e = 2(1 + 1/2! + 1/4! + ...........)
⇒ (e² + 1)/2e = 1 + 1/2! + 1/4! + ...........
⇒ (e² + 1)/2e - 1 = 1/2! + 1/4! + ...........
⇒ (e² + 1 - 2e)/2e = 1/2! + 1/4! + ...........
⇒ (e - 1)² /2e = 1/2! + 1/4! + ...........
The third term of a geometric progression isThe product of the first five terms is
0%
4
3
0%
4
5
0%
4
4
0%
none of these
Explanation
4
5
Hint:
here it is given that T
3
= 4.
⇒ ar² = 4
Now product of first five terms = a.ar.ar².ar³.ar
4
= a
5
r
10
= (ar
2
)
5
= 4
5
The sum of two numbers is 13/6 An even number of arithmetic means are being inserted between them and their sum exceeds their number byThen the number of means inserted is
0%
2
0%
4
0%
6
0%
8
Explanation
6
Hint:
Let a and b are two numbers such that
a + b = 13/6
Let A
1
, A
2
, A
3
, .........A
2n
be 2n arithmetic means between a and b
Then, A
1
+ A
2
+ A
3
+ .........+ A
2n
= 2n{(n + 1)/2}
⇒ n(a + b) = 13n/6
Given that A
1
+ A
2
+ A
3
+ .........+ A
2n
= 2n + 1
⇒ 13n/6 = 2n + 1
⇒ n = 6
If the sum of the roots of the quadratic equation ax² + bx + c = 0 is equal to the sum of the squares of their reciprocals, then a/c, b/a, c/b are in
0%
A.P.
0%
G.P.
0%
H.P.
0%
A.G.P.
Explanation
H.P.
Hint:
Given, equation is
ax² + bx + c = 0
Let p and q are the roots of this equation.
Now p+q = -b/a
and pq = c/a
Given that
p + q = 1/p² + 1/q²
⇒ p + q = (p² + q²)/(p² ×q²)
⇒ p + q = {(p + q)² - 2pq}/(pq)²
⇒ -b/a = {(-b/a)² - 2c/a}/(c/a)²
⇒ (-b/a)×(c/a)² = {b²/a² - 2c/a}
⇒ -bc²/a³ = {b² - 2ca}/a²
⇒ -bc²/a = b² - 2ca
Divide by bc on both side, we get
⇒ -c /a = b/c - 2a/b
⇒ 2a/b = b/c + c/a
⇒ b/c, a/b, c/a are in AP
⇒ c/a, a/b, b/c are in AP
⇒ 1/(c/a), 1/(a/b), 1/(b/c) are in HP
⇒ a/c, b/a, c/b are in HP
The 35th partial sum of the arithmetic sequence with terms a
n
= n/2 + 1
0%
240
0%
280
0%
330
0%
350
Explanation
350
Hint:
The 35th partial sum of this sequence is the sum of the first thirty-five terms.
The first few terms of the sequence are:
a
1
= 1/2 + 1 = 3/2
a
2
= 2/2 + 1 = 2
a
3
= 3/2 + 1 = 5/2
Here common difference d = 2 - 3/2 = 1/2
Now, a
35
= a
1
+ (35 - 1)d = 3/2 + 34 ×(1/2) = 17/2
Now, the sum = (35/2) × (3/2 + 37/2)
= (35/2) × (40/2)
= (35/2) × 20
= 35 × 10
= 350
The first term of a GP isThe sum of the third term and fifth term isThe common ratio of GP is
0%
1
0%
2
0%
3
0%
4
Explanation
3
Hint:
Let first term of the GP is a and common ratio is r.
3rd term = ar²
5th term = ar
4
Now
⇒ ar² + ar
4
= 90
⇒ a(r² + r
4
) = 90
⇒ r² + r
4
= 90
⇒ r² ×(r² + 1) = 90
⇒ r²(r² + 1) = 3² ×(3² + 1)
⇒ r = 3
So the common ratio is 3
If 2/3, k, 5/8 are in AP then the value of k is
0%
31/24
0%
31/48
0%
24/31
0%
48/31
Explanation
31/48
Hint:
Given, 2/3, k, 5/8 are in AP
⇒ 2k = 2/3 + 5/8
⇒ 2k = 31/24
⇒ k = 31/48
So, the value of k is 31/48
The sum of n terms of the series (1/1.2) + (1/2.3) + (1/3.4) + ...... is
0%
n/(n+1)
0%
1/(n+1)
0%
1/n
0%
None of these
Explanation
n/(n+1)
Hint:
Given series is:
S = (1/1·2) + (1/2·3) + (1/3·4) - ...................1/n.(n+1)
⇒ S = (1 - 1/2) + (1/2 - 1/3) + (1/3 - 1.4) -.........(1/n - 1/(n+1))
⇒ S = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 - ........... 1/n - 1/(n+1)
⇒ S = 1 - 1/(n+1)
⇒ S = (n + 1 - 1)/(n+1)
⇒ S = n/(n+1)
If the third term of an A.P. is 7 and its 7 th term is 2 more than three times of its third term, then the sum of its first 20 terms is
0%
228
0%
74
0%
740
0%
1090
Explanation
740
Hint:
Let a is the first term and d is the common difference of AP
Given the third term of an A.P. is 7 and its 7th term is 2 more than three times of its third term
⇒ a + 2d = 7 .............. 1
and
3(a + 2d) + 2 = a + 6d
⇒ 3×7 + 2 = a + 6d
⇒ 21 + 2 = a + 6d
⇒ a + 6d = 23 .............. 2
From equation 1 - 2, we get
4d = 16
⇒ d = 16/4
⇒ d = 4
From equation 1, we get
a + 2×4 = 7
⇒ a + 8 = 7
⇒ a = -1
Now, the sum of its first 20 terms
= (20/2)×{2×(-1) + (20-1)×4}
= 10×{-2 + 19×4)}
= 10×{-2 + 76)}
= 10 × 74
= 740
If the sum of the first 2n terms of the A.P. 2, 5, 8, ....., is equal to the sum of the first n terms of the A.P. 57, 59, 61, ....., then n equals
0%
10
0%
12
0%
11
0%
13
Explanation
11
Hint:
Given,
the sum of the first 2n terms of the A.P. 2, 5, 8, .....= the sum of the first n terms of the A.P. 57, 59, 61, ....
⇒ (2n/2)×{2×2 + (2n-1)3} = (n/2)×{2×57 + (n-1)2}
⇒ n×{4 + 6n - 3} = (n/2)×{114 + 2n - 2}
⇒ 6n + 1 = {2n + 112}/2
⇒ 6n + 1 = n + 56
⇒ 6n - n = 56 - 1
⇒ 5n = 55
⇒ n = 55/5
⇒ n = 11
If a is the A.M. of b and c and G
1
and G
2
are two GM between them then the sum of their cubes is
0%
abc
0%
2abc
0%
3abc
0%
4abc
Explanation
2abc
Hint:
Given, a is the A.M. of b and c
⇒ a = (b + c)
⇒ 2a = b + c ............ 1
Again, given G
1
and G
1
are two GM between b and c,
⇒ b, G
1
, G
2
, c are in the GP having common ration r, then
⇒ r = (c/b)
1/(2+1)
= (c/b)
1/3
Now,
G
1
= br = b×(c/b)
1/3
and G
1
= br = b×(c/b)
2/3
Now,
(G
1
)³ + (G
2
)3 = b³ ×(c/b) + b³ ×(c/b)²
⇒ (G
1
)³ + (G
2
)³ = b³ ×(c/b)×( 1 + c/b)
⇒ (G
1
)³ + (G
2
)³ = b³ ×(c/b)×( b + c)/b
⇒ (G
1
)³ + (G
2
)³ = b² ×c×( b + c)/b
⇒ (G
1
)³ + (G
2
)³ = b² ×c×( b + c)/b .............. 2
From equation 1
2a = b + c
⇒ 2a/b = (b + c)/b
Put value of(b + c)/b in eqaution 2, we get
(G
1
)³ + (G
2
)³ = b² × c × (2a/b)
⇒ (G
1
)³ + (G
2
)³ = b × c × 2a
⇒ (G
1
)³ + (G
2
)³ = 2abc
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