Question 1 1 / -0
Find the sum of the first 25 terms of an A.P. in which the common difference (d) is 5 and the 25th term is 250.
Solution
Given:
Common difference, d = 5
t
25 = 250
n = 25
Since t
n = a + (n - 1)d;
t
25 = a + (25 – 1) d
250 = a + (24)5
250 – 120 = a
a = 130
Thus, required sum,
s
n =
s
25 =
s
25 =
= 190 × 25 = 4750
Question 2 1 / -0
If 'p' is the A.M. of three terms of an A.P. and 'q' is the A.M. of their double, find 'p' in terms of 'q'.
Solution
Let three terms be a
1 , a
2 and a
3 .
According to the question,
p =
3p = a
1 + a
2 + a
3 --- (1)
And q =
3q = 2a
1 + 2a
2 + 2a
3 = a
1 + a
2 + a
3 = 3p (from (1))
p =
Question 3 1 / -0
In an A.P, the 17th term is 0. According to this information, which of the following statements is true?
Solution
Here, a = first term, d = common difference, Sn = Sum of n terms, n = number of terms, tn = nth termth term = 0st term = a + (31 – 1)d31 = 0 + 14d 31 = 14dst term = 14d --- (2)th term = a + (24 - 1)d24 = 0 + 7dth term = 7d --- (3)st term is double of the 24th term.
Question 4 1 / -0
Find the sum of the first 50 positive integers divisible by 9.
Solution
The positive integers divisible by 9 are: 9, 18, 27, 36, …
Here, first term a = 9, common difference d = 18 – 9 = 9
The 50
th term is:
a
50 = 9 + (50 – 1) × 9 = 9 + 441 = 450
So, the sum of the first 50 terms is:
S
50 =
=
= 25 × 459 = 11,475
Question 5 1 / -0
The first term of an arithmetic progression is 9 times of its 9th term. If the common difference is -2, then which of the following will represent the 6th term?
Solution
Suppose, first term = a, and common difference = d = -2th term;th term = 18 + (5 × (-2))
Question 6 1 / -0
The sum of three consecutive terms of an A.P. is 81 and their product is 19,440. Find out these terms.
Solution
Let the consecutive terms be (a - d), a, (a + d).2 - d2 = 19,440 ÷ 272 = 720 2 = 720 - 729 2 = 9
Question 7 1 / -0
The sum of the first 10 terms and sum of the first 20 terms of an A.P. are 14,520 and 2,68,100, respectively. Find the first term and the common difference.
Solution
Here, n = number of terms, a = first term, d = common difference, S
n = sum of n terms, t
n = n
th term
Sum of 10 terms =
(2a + (n – 1)d)
14,520 =
(2a + (10 - 1)d)
14,520 = 5(2a + 9d)
2904 = 2a + 9d --- (1)
Sum of 20 terms =
(2a + (n - 1)d)
2,68,100 =
(2a + (20 - 1) d)
26,810 = 2a + 19d --- (2)
Subtracting equation (1) from (2),
26,810 - 2904 = 2a + 19d - (2a + 9d)
23,906 = 10d
d = 2390.6
Put this value in equation (1).
2904 = 2a + 9(2390.6)
2904 = 2a + 21515.4
2904 - 21515.4 = 2a
a = -9305.7
Question 8 1 / -0
In an A.P, the 51st term is 5 times the 15th term. According to this information, which of the following statements is true?
Solution
Here, a = first term, d = common difference st term = 5(15th term)
Question 9 1 / -0
The first term of an A.P. is 3, the second term is 0 and the third term is -3. If -27 is the last term of the A.P, then what is the sum of all these terms?
Solution
Here, a = first term, d = common difference, S
n = sum of n terms, t
n = n
th term
The series is 3, 0, -3, ..., -27.
Here, a = 3 and d = -3
Therefore,
-27 = 3 + (n – 1)(-3)
= n – 1
10 + 1 = n
n = 11
S
11 =
[3 + (-27)]
=
× (-24)
= -132
Question 10 1 / -0
In an A.P, the 16th term is -19. If the sum of the first 11 terms is 77, then which of the following values represents the sum of the first 20 terms of the A.P?
Solution
Here, a = first term, d = common difference, S
n = sum of n number of terms, n = number of terms
16
th term = -19
a + (16 – 1)d = -19
a + 15d = -19 --- (1)
Sum of the first 11 terms = 77
(2a + (n – 1)d) = 77
(11/2)(2a + 10d) = 77
2a + 10d = 14 --- (2)
Multiplying equation (1) by 2;
2(a + 15d = -19)
2a + 30d = -38 --- (3)
Subtracting equation (3) from (2);
2a + 10d – 2a – 30d = 14 - (-38)
-20d = 52
d = -2.6
Use the value of d in equation (1).
a + 15(-2.6) = -19
a – 39 = -19
a = -19 + 39 = 20
Sum of the first 20 terms =
(2a + (n – 1)d)
=
(2 × 20 + 19 × - 2.6)
= 10(40 – 49.4) = - 94
Question 11 1 / -0
The ratio of sum of n terms of two arithmetic progressions is (2n + 3) : (4n + 5). Find the ratio of their 10th term.
Solution
Ratio of sum of n terms =
Or
=
Or
=
When n = 19
=
Or
=
Thus,
is the required ratio.
Question 12 1 / -0
Find the sum of all integers between 50 and 500 which are divisible by 7.
Solution
a
1 = 56 = 7 × 8
a
n = 497 = 71 × 7
So, S
n = 56 + 63 + 70 + ............ + 497
= 7(8 + 9 + 10 + ........... + 71)
Numbers of terms = 71 - 8 + 1 = 64
S
n = (
) × 7 × (8 + 71)
= 7 × 79 × 32 = 17696
Question 13 1 / -0
The sum of 3rd and 15th terms of an arithmetic progression is equal to the sum of 6th , 11th and 13th terms of the same progression. Which term of the given progression should necessarily be equal to zero?
Solution
Let the AP be a, a + d, a + 2d, …
T
n = a + (n - 1)d
T
3 + T
15 = T
6 + T
11 + T
13 2a + 2d + 14d = 3a + 5d + 10d + 12d
0 = a + 11d
0 = T
12 Thus, the 12
th term of the AP is 0.
Question 14 1 / -0
If mth and nth terms of an A.P. are n and m respectively, then the (m + n)th term is
Solution
a + (m - 1) d = n ............(1)m + n = a + (m + n - 1) d = m + n - 1 + (m + n - 1) (- 1) = 0
Question 15 1 / -0
In an A.P, the common difference is 11. If the sum of the first five terms is -85, then which of the following values represents the value of the second term?
Solution
Here, a = first term, d = common difference, S
n = sum of n number of terms, t
n = n
th term
n = 5, S
5 = –85 and d = 11
S
n =
(2a + (n - 1)d)
Therefore, –85 =
(2a + 4 × 11)
× 2 = 2a + 44
–34 – 44 = 2a
–78 = 2a
a = –39
t
2 = a + (2 - 1)d
= a + d = -39 + 11
= -28
Question 16 1 / -0
There is a lemon race in a school. A lemon holder is placed at the starting point, which is 7 m away from the first lemon, and other lemons are placed 4 m apart in a straight line. Ollie starts from the lemon holder, picks up the first lemon and runs back to the holder to place the first lemon, and then runs backs to pick the next lemon. She continues in the same way until all the lemons are in the holder. If there are 20 lemons, then what is the total distance covered by Ollie?
Solution
The distances of lemons from the holder are 7 m, 11 m, 15 m, 19 m, ....
The distances run by Ollie for collecting these lemons are two times the distance at which the lemons have been kept.
Therefore, distances to be run are
14, 22, 30, 38........
First term (a) = 14
Common difference (d) = 22 - 14 = 8
S
20 =
(2a + (n - 1)d)
=
([2 × 14] + (20 - 1)8)
= 10(28 + 152)
= 1800 m
Question 17 1 / -0
Molly is playing with some sticks. She wants to stack them. So, she places 20 sticks in the bottom row, 19 in the next row, 18 in the next, and so on. If she only has 200 sticks, then how many sticks will be there in the topmost row?
Solution
Numbers of sticks in the rows are given below:
20, 19, 18, 17, ...
For this A.P,
First term = a = 20
Common difference = d = 19 - 20 = -1
Suppose, 200 sticks are placed in n rows.
S
n = 200
S
n =
(2a + (n - 1)d)
200 =
( [2 × 20] + (n - 1)(-1))
200 =
(40 - n + 1)
400 = 40n - n
2 + n
n
2 - 41n + 400 = 0
n
2 - 16n - 25n + 400 = 0
n(n - 16) -25(n - 16) = 0
(n - 25)(n - 16) = 0
n = 25 or 16
a
n = a + (n - 1)d
a
16 = 20 + (16 - 1)(-1)
= 20 - 15 = 5
a
25 = 20 + (25 - 1)(-1)
= 20 - 24
= -4
The number of sticks cannot be negative.
So, the number of sticks in the topmost row is 5.
Question 18 1 / -0
A scientist is observing a galaxy. It is in spiral shape which is made up of semicircles. The radius of the first semicircle is
billion kilometres, the radius of the second semicircle is 2 billion kilometres, the radius of the third semicircle is
billion kilometres, and so on. According to this information, which of the following statements is true?
Solution
Perimeter of a semicircle = πr
p
1 = π(
) =
billion km
p
2 = π(2) = 2π billion km
p
3 = π(
) =
billion km
Here, p
1 , p
2 and p
3 are the perimeters of the 1
st , 2
nd and 3
rd semicircles.
First term (a) =
Common difference(d) = 2π -
=
S
n =
(2a + (n - 1)d)
S
121 =
([2 ×
] + (121 - 1)
)
=
(3π + 60π)
=
× 63 ×
= 11979 billion km
Question 19 1 / -0
A company announces some cash prize to celebrate the hard work of its employees. It decides a sum of Rs. 50,000 to be used to give 5 prizes to the selected employees for their overall performance. Each prize is Rs. 4000 less than its preceding prize. According to this information, which of the following represents the value of the third prize?
Solution
Let the value of the first prize be p.
Value of the second prize = p - 4000
Value of the third prize = p - 8000
So, first term = a = p
Common difference = d = (p - 8000) - (p - 4000) = -4000
S
5 = 50,000
50,000 =
( {2 × p} +(5 - 1)(-4000))
20,000 = 2p + (-16,000)
20,000 = 2p - 16,000
p = 18,000
Therefore, the values will be Rs. 18,000, Rs. 14,000, Rs. 10,000, Rs. 6000 and Rs. 2000.
So, the value of the third prize will be Rs. 10,000.
Question 20 1 / -0
In a field, there are few plants in rows. There are 80 plants in the first row, 76 plants in the second row, 72 plants in the third row, and so on. There are only 24 plants in the last row. How many rows are there in the field?
Solution
Here, a = first term, d = common difference, Sn = sum of n number of terms, tn = nth termn = a + (n - 1)d
Question 21 1 / -0
Which of the following statements are true?
Solution
Here, a = first term, d = common difference, S
n = sum of n number of terms, t
n = n
th term
(1) The series is 1, 3, 5, ..., 197.
Here, a = 1 and d = 3 – 1 = 2
So, 197 = a + (n – 1)d
197 = 1 + (n – 1)2
= n – 1
98 + 1 = n
n = 99
Hence, S =
=
= 99 × 99
= 9801
(2) First multiple of 11 = 110
Last multiple of 11 = 2321
So, a = 110, l = 2321, d = 11
l = a + (n - 1)d
2321 = 110 + (n - 1)11
2321 - 110 = 11n - 11
2211 + 11 = 11n
2222 = 11n
202 = n
(3) a + 1, 3a, 4a + 2 are in A.P.
If a, b, c are in A.P, then 2b = a + c.
Therefore, 2(3a) = a + 1 + 4a + 2
6a = 5a + 3
6a - 5a = 3
a = 3
Question 22 1 / -0
In a certain A.P, the first term, the common difference and the nth term are represented by x, y and z. According to this information, which of the following represents the value of n?
Solution
Here, a = first term, d = common difference, S
n = sum of n number of terms, a
n = n
th term
Now, a = x, d = y
a
n = z
a
n = a + (n – 1)d
z = x + (n – 1)y
z – x = ny – y
n =
Question 23 1 / -0
A certain A.P. is given and the A.P. is -200, -170, -140, -110, ...
Solution
Here, a = first term, d = common difference, S
n = sum of n terms, n = number of terms, t
n = n
th term
d = -170 - (-200) = -170 + 200 = 30, n = 50
Sum of n terms =
(2a + (n – 1)d)
=
([2 × -200] + (50 - 1) × 30)
= 25(-400 + 49 × 30)
= 25(-400 + 1470)
= 25(1070)
S
50 = 26,750
n = 20
Sum of n terms =
(2a + (n – 1)d)
= (20/2)([2 × -200] + (20 -1) × 30)
= 10(-400 + 19 × 30)
= 10(-400 + 570)
= 10(170)
S
2 0 = 1700
Question 24 1 / -0
There are two arithmetic progressions. The first, the second and the third term of the first A.P. are 28, 26.5 and 25, respectively; and the first three terms of the second A.P. are 4, 4.5 and 5, respectively. If the nth terms of the two A.P.s are equal, then which of the following values represents 'n'?
Solution
Here, a = first term, d = common difference, Sn = sum of n number of terms, tn = nth term1 = 28, 26.5, 25 1 = 281 = 26.5 - 28 = -1.52 = 4, 4.5, 5, ...2 = 5 2 = 4.5 - 4 = 0.5th terms of these A.P.s are equal.1 + (n - 1)d1 = a2 + (n - 1)d2
Question 25 1 / -0
Fill in the blanks.th term exceeds the 21st term by 12, then the common difference of the A.P. P . Q . R .
Solution
Here, first term = a, common difference = d, t
n = n
th term, n = number of terms, S
n = sum of n number of terms
(1) t
5 = 14, t
27 = t
21 + 12
14 = a + (n - 1)d
14 = a + 4d
a = 14 - 4d ... (1)
According to the question,
t
27 = t
21 + 12
a + (27 - 1)d = a + (21 - 1)d + 12
a + 26d = a + 20d + 12
From equation (1),
14 - 4d + 26d = 14 - 4d + 20d + 12
14 + 22d = 26 + 16d
22d - 16d = 26 - 14
6d = 12
d = 2
(2) a = a, S
n = c and l = b
c =
[a + b]
2c = n(a + b)
n =
(3) Amount of money donated = (Class × 100) × 4 (4 = Number of sections)
Class I = (1 × 100) × 4 = 400
Class II = (2 × 100) × 4 = 800
Class III = (3 × 100) × 4 = 1200
It forms an A.P, where
First term (a) = 400
Common difference (d) = 800 - 400 = 400
Number of terms (n) = 12
S
n =
(2a + (n - 1)d)
S
12 =
([2 × 400] + (12 - 1)400)
= 6(800 + 4400)
= Rs. 31,200
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