Series Test

Result

Series Test - 5

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Weekly Quiz Competition

Question 1
1 / -0

The 6^th and 17^th terms of an A.P. are 19 and 41 respectively. Then the 40^th term is

A
87

B
80

C
60

D
100

Solution

a + 5d = 19 ……(i)
a + 16d = 41 ….… (ii)
Subtracting (i) from (ii), we get
11d = 22
⇒ d = 2
From (1), a + 5 × 2 = 19
⇒ a = 9
Now 40^th term = a + 39d
= 9 + 39 × 2
= 9 + 78 = 87
Question 2
1 / -0

Find the maximum value of the sum of the following series:

44, 40, 36, 32, ...

A
220

B
264

C
232

D
320

Solution

As the difference between 2 consecutive terms is the same, this is an AP.
1^st term of this AP = a = 44
Common difference = d = - 4
Now, the sum will be maximum, if all the terms are positive.
Let the n^th term be the smallest positive term of this AP.
Thus, T_n = a + (n - 1)d = 44 + (n - 1)(- 4) ≥ 0
Or 44 + (n - 1)(- 4) ≥ 0
Or 48 - 4n ≥ 0
Thus, 12 ≥ n
Thus, for maximum sum, n = 12
S_n= (2a + (n - 1)d)
Putting the values n = 12, a = 44, and d = - 4 in the above equation, we get
S₁₂ = 264
Thus, answer option 2 is correct.
Question 3
1 / -0

The sums of n terms of two A.P. are in the ratio . The ratio of their sixth term is

A
32 : 59

B
1 : 1

C
2 : 1

D
53 : 119

Solution

=
=
Putting n = 11 in R.H.S, we get
= = , which is equal to ratio of 6^th term.
Now, putting n = 11, =
Question 4
1 / -0

There are four terms in an A.P. The sum of the two means is 16 and the product of the two extremes is 28.
Find the greatest term.

A
6

B
10

C
8

D
14

Solution

Let 4 terms be a, b, c, d.
Sum of extremes = Sum of means
So, a + d = b + c = 16
And ad = 28
So, a = 14, d = 2
Or d = 14, a = 2
Question 5
1 / -0

If the p^th, q^thand r^th terms of an A.P. are x, y and z respectively, then x(q - r) + y(r - p) + z(p - q) is

A
- 1

B
0

C
1

D
3

Solution

a + (p - 1)d = x … (i)
a + (q - 1)d = y …(ii)
a + (r - 1)d = z …(iii)
Subtracting (ii) from (i), (iii) from (ii) and (i) from (iii), we get
(p - q)d = x - y (p - q) =
(q - r)d = y - z (q - r) =
(r - p)d = z - x (r - p) =
Given, x(q - r) + y(r - p) + z(p - q)
= + + = = = 0
Question 6
1 / -0

If m^th and n^th terms of an A.P. are n and m respectively, then the (m + n)^th term is

A
0

B
1

C
- 1

D
m + n

Solution

a + (m - 1) d = n ............(1)
a + (n - 1) d = m ............(2)
By subtracting (2) from (1), we get (m - n) d = (n - m)
d = - 1 .............(3)
By substituting the value of d from equation (3) into equation (1), we get a + (m - 1)( - 1) = n
a = n + m - 1
T_{m + n} = a + (m + n - 1) d = m + n - 1 + (m + n - 1) (- 1) = 0
Question 7
1 / -0

Find the sum of all the positive terms of A.P. 19, 18, 17, ….

A
230

B
234

C
234

D
235

Solution

The series can be written as .
The last term is because after that the terms of series become negative.
S = [ 95 + 91 + 87 + ………+ 3]
The number of terms in the series,
a_n = a + (n - 1)d
3 = 95 + (n - 1)(-4)
n = 24
Thus, S = [ (95 + 3) ] {S = (a + l)}
= 235
Question 8
1 / -0

The nth term of the series is 2n - 5. Find the ratio of the sum of the first five terms to the next five terms.

A
1 : 11

B
2 : 5

C
1 : 2

D
3 : 8

Solution

T_n = 2n - 5, Putting n = 1, 2, 3 â¦â¦, the series is - 3, - 1, 1, 3, 5, 7, 9, 11, 13, 15
Thus, the sum of the first five terms to the sum of the next five terms = 5/55 or 1 : 11.
Question 9
1 / -0

The ratio of sum of n terms of two arithmetic progressions is (2n + 3) : (4n + 5). Find the ratio of their 10^th term.

A

B

C

D

Solution

Ratio of sum of n terms = Or = Or =
When n = 19
=
Or =
Thus, is the required ratio.
Question 10
1 / -0

If k +2, 4k - 6 and 3k - 2 are three consecutive terms of an A.P., then k =

A
0

B
1

C
2

D
3

Solution

Common difference = (4K - 6) - (K + 2) = (3K - 2) - (4K - 6)
3K - 8 = - K + 4
4K = 12
K = 3
Question 11
1 / -0

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

A
0

B
p + q + r

C
p + q - r

D
p - q - r

Solution

a + (p - 1) d = q
& a + (q - 1) d = p
d = -1 & a = p + q - 1
T_r = a + (r - 1) d = p + r - 1 + (r - 1) - 1 = p + q - r
Question 12
1 / -0

If x, y, z are in A.P., find the value of (x + 2y - z)(2y + z - x)(z + x -y).

A
xyz

B
2xyz

C
4xyz

D
None of these

Solution

As x, y, z are in A.P, therefore 2y = x + z
The given expression is (x + 2y – z)(2y + z – x)(z + x – y).
= (x + x + z – z) (x + z + z – x) (2y – y)
= 2x ´ 2z ´ y = 4xyz
Question 13
1 / -0

If a, b, c and d are in G.P., then (a - c)² + (b -c)² + (b - d)² is equal to

A
0

B
(a - d)²

C
(b - d)²

D
1

Solution

a, b, c, d are in G.P.
b² = ac, c² = bd and ad = bc
Given equation: a² + c² - 2ac + b² + c² - 2bc + b² + d² - 2bd
= [a² + 2b² + 2c² + d² - 2ac - 2bc - 2bd]
= [a² + d² + 2ac + 2bd - 2ac - 2bc - 2bd]
= [a² + d² - 2ad] = (a - d)²
Question 14
1 / -0

If the 5^th, 8^th and 11^thterms of a G.P. are p, q and s respectively, then

A
q = ps

B
q² = ps

C
q³= ps

D
None of these

Solution

ar⁴ = p, ar⁷ = q and ar¹⁰ = s
ps = (ar⁴) (ar¹⁰) = (a² r¹⁴) = (ar⁷)² = q²
Question 15
1 / -0

+ + +……

A

B
3

C
³

D
None of these

Solution

+ + +…
a = , r =
S= = = =
Question 16
1 / -0

If x, 2x + 2, 3x + 3 are in a GP, then the fourth term is

A
– 27

B
27

C
– 13.5

D
13.5

Solution

x, 2x + 2, 3x + 3 are in a GP.
(2x + 2)² = 3(x + 1) x
4 (x + 1) = 3x
⇒ x = – 4
So, the terms are – 4, – 6, – 9.
a = – 4 and r =
Fourth term = ar³⇒ – = – 13.5
Question 17
1 / -0

If the sum of the series to is a finite number, then

A
x < 3

B
x >

C
x <

D
x > 3

Solution

Since the sum of ¥ of the given GP exists, therefore its common ratio should be less than unity, i.e. < 1
Or x > 3
Question 18
1 / -0

If x, y, z are in A.P., then , and are in

A
A.P.

B
G.P.

C
H.P.

D
A.G.P.

Solution

x, y, z are in A.P.
2y = (z + x) [Add (z + x) on both sides.]
2y + (z + x) = 2(z + x)
2(z + x) = (y + z) + (x + y)
(y + z), (z + x), (x + y) are in A.P.
, , are in H.P.
Question 19
1 / -0

If , , are in HP, then x, y, z, are in

A
AP

B
GP

C
HP

D
None of these

Solution

, , are in HP
x + y, 2y, y + z are in AP
4y = x + y + y + z = 2y + x + z
2y = x + z
x, y, z are in AP
Question 20
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.