Series Test

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Series Test - 6

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Weekly Quiz Competition

Question 1
1 / -0

If a₁, a₂, a₃ are in H.P., then

A
a₁a₂ + a₂a₃ + a₃a₁ = 0

B
a₁a₂ + a₂a₃ = a₃a₁

C
a₁a₂ + a₂a₃ = 2a₁a₃

D
a₁a₂ + a₂a₃ + 2a₁a₃ = 0

Solution

If a₁, a₂, a₃ are in H.P.
then , , are in A.P.
2 = +
2a₁a₃ = a₂a₃ + a₁a₂
Question 2
1 / -0

If p^thterm of an H.P is qr and q^thterm is pr, then r^th term of the H.P is

A
pqr

B
1

C
pq

D
pqr²

Solution

P^th term = = qr
⇒ a + (p – 1) d = … (1)
Similarly a + (q – 1) d = … (2)
Subtracting (2) from (1), we get
(p –q) d = – = d =
Substituting the value of d in (1)
a + (p – 1) =
a = – = =
R^th term = = = = pq.
Question 3
1 / -0

If a, b, c are in H.P., then the value of is

A

B

C

D
All the above

Solution

As a, b, c are in H.P. 1/a, 1/b, 1/c are in A.P. or â¦ (A)
Therefore [using (A)]
= =
Also =
Lastly = = .
Question 4
1 / -0

If the ratio of harmonic mean between two positive numbers to their geometric mean is 12 : 13, then the numbers are in the ratio

A
12 : 13

B
1/12 : 1/13

C
1 : 1

D
4 : 9

Solution

Let the two numbers be x and y.
=
=
or 2 13 = 12 (x + y)
Now check by options, = satisfies the above equation.
Question 5
1 / -0

If H is H.M. between a and b, then the value of H/a + H/b is

A
2

B

C

D
1

Solution

Harmonic mean of a and b = H = Ã + = H = = 2
Question 6
1 / -0

Which of the following terms of the A.P. must be zero if the sum of 3^rd and 7^th term of the A.P. is equal to 4^th term of the A.P.?

A
9^th

B
8^th

C
7^th

D
6^th

Solution

T₃ + T₇ = T₄a + 2d + a + 6d = a + 3d
2a + 8d = a + 3d
a + 5d = 0
T₆ = 0
Option 4 is correct.
Question 7
1 / -0

If is the A.M. between a and b, then the value of n is

A
0

B
1

C
- 1

D
None of these

Solution

According to the question,

(a + b) (a^{n - 1} + b^{n - 1}) = 2aⁿ + 2bⁿ
aⁿ + bⁿ + ba^{n - 1} + ab^{n - 1} = 2aⁿ+ 2bⁿ
ba^{n - 1} + ab^{n - 1} = aⁿ + bⁿ
aⁿ = aⁿ + bⁿ
+ = 1 +
n = 1
Option 2 is correct.
Question 8
1 / -0

Find four numbers in a GP whose sum is 85 and product is 4096.

A
1, 4, 16, 64

B
2, 3, 18, 62

C
1, 6, 15, 65

D
None of these

Solution

Only in option (1), numbers are in a GP whose sum is 85 and product is 4096.
Question 9
1 / -0

The sum of three numbers in a GP is 14. If 1 is added to the first and the second terms and 1 is subtracted from the third term, the new numbers are in AP. The smallest of them is

A
2

B
4

C
6

D
8

Solution

a + ar + ar² = 14 ....(i)
Now, a + 1, ar + 1, ar² –1 are in AP.
2(ar + 1) = a + 1 + ar² – 1
2ar + 2 = a + ar²....(ii)
From (i) and (ii), 2ar + 2 = 14 – ar
3ar = 12
ar = 4 …..(iii)
From (i), a + 4 + 4r = 14
a + 4r = 10 … (iv)
From (iii) and (iv), a + 16/a = 10
a² - 10a + 16 = 0
a² -8a - 2a + 16 = 0
(a - 2)(a - 8) = 0
a = 2, 8
Therefore, the smallest number is 2.
Question 10
1 / -0

The 288^th term of the series abbcccddddeeeeefffffffâ¦. is

A
u

B
v

C
w

D
x

Solution

The series a, b, b, c, c, c, d, d, d, d, e, e, e, e, eâ¦
The series is 1, 2, 3, 4, 5 and so on.
Sum of n integers, starting from 1, is given by:

Now, < 288
n₁ (n₁ + 1) < 576
If n₁ = 24, L.H.S < 576
Thus, for n₁ = 23
= = 23 12 = 276
Thus, n₁ = 24^th letter will start the series from 277^th term.
Also, n₁ = 24 corresponds to `x`.
Question 11
1 / -0

The sum of the series, 2 + 12 + 36 + 80 + 150 ______ up to 30 terms is

A
265280

B
262680

C
248880

D
225680

Solution

General term, t_n = n³ + n²
Sum of n terms, S_n = = +
S_n = [n(n + 1)/2]² + [n(n + 1)(2n + 1)/6]
Putting n = 30, we get the sum as 225680.
Question 12
1 / -0

The sum of 20 terms of the series, ……….. is equal to

A

B

C

D

Solution

……….. 20 term

=
=
= = =
Or is the sum of 20 terms.
Question 13
1 / -0

5 + 55 + 555 + ... up to the n^th term is equal to

A
(5/81) [10^{n + 1} - 9n - 10]

B
(5/81) [10ⁿ - 9n + 10]

C
(5/9) [10ⁿ - 9n + 10]

D
None of these

Solution

Given series: 5 + 55 + 555 … n terms
= 5 (1 + 11 + 111 + 1111 …n terms)
= (9 + 99 + 999 … n terms)
= ((10 – 1) + (100 – 1) … n terms)
= ((10 + 100 …n terms) – n)
=
= (10^{n + 1} – 9n – 10)
Question 14
1 / -0

Find the sum upto the n^th term of the series 0.7 + 0.77 + 0.777 + _____.

A
[10ⁿ- 1 - 9n]

B
[n - [1 - (0.1)ⁿ]/9]

C
[10n - 1 - 9n]

D
None of these

Solution

0 .7 + 0.77 + 0.777--------- + n^th term= + ---- + n^thterm
=
=
=
=
= = =
Question 15
1 / -0

The sum of n terms of the series 1 + (1 + 3) + (1 + 3 + 5) + ... is

A
n²

B

C

D
None of these

Solution

n^thterm of the given series is a_n = 1 + 3 + 5 + … + (2n – 1) = n²
Therefore, the sum of n terms of the series = = =
Question 16
1 / -0

If the sum of an infinite GP is 3 and the sum of the squares of its terms is also 3, then its first term and the common ratio, respectively, are

A
3/2, 1/2

B
1/2, 3/2

C
1, 1/2

D
None of these

Solution

Let 'a' be the first term and 'r' be the common ratio.

Then = 3 … (1)
a = 3 (1 - r)
It is given that a² + a² r² + a² r⁴ + ………. = 3
= 3 … (2)
Put the value of a in equation (2),
= 3 3 (1 – r) = 1 + r r =
From (1), the value of a = 3/2
Question 17
1 / -0

If a, b, c are in A.P., b - a, c - b and a are in G.P., then a: b: c is

A
1 : 2 : 3

B
1 : 3 : 5

C
2 : 3 : 4

D
1 : 2 : 4

Solution

2b = a + c (given)
(c - b)² = (b - a) a (given)
(b - a)² = (b - a) a (using c - b = a - b)
b = 2a c = 3a (using 2b = a + c).
Therefore a : b : c = 1 : 2 : 3
Question 18
1 / -0

The sum of infinite series 1 + 2 + 3 + … is

A
n²

B
n (n + 1)

C
n

D
None of these

Solution

Let x = 1 - , and let
S = 1 + 2x + 3x² + 4x³ + …
Sx = x + 2 x² + 3 x³ + …
S (1 - x) = 1 + x + x² + x³ + …
S (1 - x) =
S =
S = = n².
Question 19
1 / -0

The interior angles of a polygon are in A.P. The smallest angle is 120^o and the common difference between the angles is 5^o. Find the number of sides of the polygon.

A
6

B
7

C
8

D
9

Solution

Let n be the number of sides of the polygon. Thus, the sum of its interior angles is given by
S_n = (n - 2) 180°. ……(i)
Hence, the interior angle forms an A.P. whose first term a = 120° and common difference = 5°.
∴ S_n = [2 × 120° + (n - 1) × 5°) = [240° + (n - 1)5°] ……(ii)
By equating (i) and (ii), we get
(n - 2) × 180° = [240° + (n - 1) 5°]
(n - 2) × 360 = n (240 + 5n - 5) = n (235 + 5n)
360n - 720 = 235n + 5n²
5n² - 125n + 720 = 0
or n² - 25n + 144 = 0 or (n - 16) (n - 9) = 0
n = 16 or n = 9
But when n = 16, the last angle = T_n = a + (n - 1) d = 120° + 15(5°) = 120° + 75°
= 195°, which is not possible [Angle > 180°]. Hence, n = 9
Question 20
1 / -0

If the sum of the first 11 terms of an arithmetic progression is equal to the sum of the first 19 terms, then what is the sum of the first 30 terms?

A
48

B
36

C
24

D
0

Solution

a, a + d, a + 2d, a + 3d, ........... in AP.
Sum of 11 terms = 11a + d
= 11a + 55d
Sum of the first 19 terms = 19a + d
= 19a + 171d
According to the question,
11a + 55d = 19a + 171d
55d - 171d = 8a
= a
= a
Or a + = 0
Or 2a + 29d = 0
Sum of the first 30 terms = 30a + 29 d
Or 30a + 29 15d
Or 15[2a + 29d] = 0
So, the sum of 30 terms is equal to zero.