CBSE Class 10 Arithmetic Progressions Sure Shot Questions Set 05

Question. Find the \( 9^{th} \) term from the end (towards the first term) of the AP: 5, 9, 13, \dots, 185.
Answer: 153

Question. For what value of \( k \) will \( k + 9, 2k - 1 \) and \( 2k + 7 \) are the consecutive terms of an AP?
Answer: 18

Question. For what value of \( k \) will the consecutive terms \( 2k + 1, 3k + 3 \) and \( 5k - 1 \) form an AP?
Answer: 6

Question. Find the eleventh term from the last term of the AP: 27, 23, 19, \dots, \( - 65 \).
Answer: \( - 25 \)

Question. If the first three terms of an AP are \( b, c \) and \( 2b \), then find the ratio of \( b \) and \( c \).
Answer: \( 2 : 3 \)

Question. Find the value of \( x \) so that \( - 6, x, 8 \) are in AP.
Answer: 1

Question. Find the \( 11^{th} \) term of the AP: \( - 27, - 22, - 17, - 12, \dots \).
Answer: 23

Question. The \( n^{th} \) term of an AP is \( (7 - 4n) \), then what is its common difference?
Answer: - 4

Question. Find the common difference of the AP whose first term is 12 and fifth term is 0.
Answer: - 3

Question. Find how many integers between 200 and 500 are divisible by 8.
Answer: 37

Question. Which term of the progression \( 20, 19\frac{1}{4}, 18\frac{1}{2}, 17\frac{3}{4}, \dots \) is the first negative term?
Answer: 28

Question. Is \( - 150 \) a term of the AP: 17, 12, 7, 2, \dots?
Answer: No

Question. Find the number of two-digit numbers which are divisible by 6.
Answer: 15

Question. Which term of the AP: 3, 14, 25, 36, \dots will be 99 more than its \( 25^{th} \) term?
Answer: 34

Question. Which term of the AP: 3, 15, 27, 39, \dots will be 120 more than its \( 21^{st} \) term?
Answer: 31

Question. How many natural numbers are there between 200 and 500, which are divisible by 7?
Answer: 43

Question. How many two-digit numbers are divisible by 7?
Answer: 13

Question. How many two digits numbers are divisible by 3?
Answer: 30

Question. If \( \frac{1}{x + 2}, \frac{1}{x + 3} \) and \( \frac{1}{x + 5} \) are in AP, find the value of \( x \).
Answer: 1

Question. How many three digit numbers are divisible by 11?
Answer: 81

Question. In an AP, the first term is 12 and the common difference is 6. If the last term of the AP is 252, find its middle term.
Answer: 132

Question. Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5.
Answer: 89

Question. The \( 4^{th} \) term of an AP is zero. Prove that the \( 25^{th} \) term of the AP is three times its \( 11^{th} \) term.
Answer: Let the first term be \( a \) and common difference be \( d \).
Given, \( a_4 = 0 \Rightarrow a + (4-1)d = 0 \Rightarrow a = -3d \).
Now, \( a_{25} = a + (25-1)d = -3d + 24d = 21d \).
Also, \( a_{11} = a + (11-1)d = -3d + 10d = 7d \).
Clearly, \( 21d = 3(7d) \Rightarrow a_{25} = 3 \times a_{11} \). Hence proved.

Question. Find the middle term of the AP: 6, 13, 20, \dots, 216.
Answer: 111

Question. The \( n^{th} \) term of an AP is \( 6n + 2 \). Find its common difference.
Answer: 6

Question. Find the \( 10^{th} \) term from end of the AP: 4, 9, 14, \dots, 254.
Answer: 209

Question. Determine \( k \) so that \( k^2 + 4k + 8, 2k^2 + 3k + 6, 3k^2 + 4k + 4 \) are three consecutive terms of an AP.
Answer: 0

Question. Find the number of natural numbers between 102 and 998 which are divisible by 2 and 5 both.
Answer: 89

Question. Which term of the AP: 115, 110, 105, \dots is its first negative term?
Answer: \( 25^{th} \) term

Question. If the \( 9^{th} \) term of an AP is zero, prove that its \( 29^{th} \) term is double of its \( 19^{th} \) term.
Answer: Let first term be \( a \) and common difference be \( d \).
Given \( a_9 = 0 \Rightarrow a + 8d = 0 \Rightarrow a = -8d \).
Now, \( a_{29} = a + 28d = -8d + 28d = 20d \).
Also, \( a_{19} = a + 18d = -8d + 18d = 10d \).
Clearly, \( 20d = 2(10d) \Rightarrow a_{29} = 2 \times a_{19} \). Hence proved.

Question. The angles of a triangle are in AP. The greatest angle is twice the least. Find all the angles of the triangle.
Answer: 40°, 60°, 80°

Question. For what value of \( n \), the \( n^{th} \) term of two APs: 63, 65, 67, \dots and 3, 10, 17, \dots are equal.
Answer: 13

Question. The \( 8^{th} \) term of an AP is 37 and its \( 12^{th} \) term is 57. Find the AP.
Answer: 2, 7, 12, 17, 22, \dots

Question. The \( p^{th}, q^{th} \) and \( r^{th} \) terms of an AP are \( a, b \) and \( c \) respectively. Show that \( a(q - r) + b(r - p) + c(p - q) = 0 \).
Answer: Let \( A \) and \( d \) be the first term and common difference of the given AP, then
\( a_p = A + (p - 1)d = a \dots(i) \)
\( a_q = A + (q - 1)d = b \dots(ii) \)
\( a_r = A + (r - 1)d = c \dots(iii) \)
Now, subtracting (i) and (ii), we get \( (p - q)d = a - b \Rightarrow p - q = \frac{a - b}{d} \).
Multiplying by ‘\( c \)’ on both sides, \( c(p - q) = \frac{ca}{d} - \frac{cb}{d} \dots(iv) \).
Now, (ii) - (iii), we get \( (q - r)d = b - c \Rightarrow q - r = \frac{b - c}{d} \).
Multiplying by ‘\( a \)’ on both sides, \( a(q - r) = \frac{ab}{d} - \frac{ac}{d} \dots(v) \).
Now, (iii) - (i), we get \( (r - p)d = c - a \Rightarrow r - p = \frac{c - a}{d} \).
Multiplying by ‘\( b \)’ on both sides, \( b(r - p) = \frac{bc}{d} - \frac{ba}{d} \dots(vi) \).
Adding (iv), (v) and (vi), we get \( a(q - r) + b(r - p) + c(p - q) \)
\( = \frac{ab}{d} - \frac{ac}{d} + \frac{bc}{d} - \frac{ba}{d} + \frac{ca}{d} - \frac{cb}{d} = 0 \). Hence proved.

Question. If the \( n^{th} \) terms of two APs: 23, 25, 27, \dots and 5, 8, 11, 14, \dots are equal, then find the value of \( n \).
Answer: 19

Question. If \( m \) times the \( m^{th} \) term of an Arithmetic Progression is equal to \( n \) times its \( n^{th} \) term and \( m \neq n \), show that the \( (m + n)^{th} \) term of the AP is zero.
Answer: We know that \( a_n = a + (n - 1)d \)
From the given conditions,
\( m[a + (m - 1) d] = n[a + (n - 1)d] \)
\( \Rightarrow m[a + (md - d)] = n[a + nd - d] \)
\( \Rightarrow am + m^2d - md = an + n^2d - nd \)
\( \Rightarrow am - an + m^2d - n^2d - md + nd = 0 \)
\( \Rightarrow a(m - n) + d(m^2 - n^2) - d(m - n) = 0 \)
\( \Rightarrow a(m - n) + (m + n) (m - n)d - (m - n)d = 0 \)
\( \Rightarrow (m - n) [a + (m + n) d - d] = 0 \)
\( \Rightarrow a + md + nd - d = 0 \)
\( \Rightarrow a + (m + n - 1)d = 0 \)
Since, \( m \neq n \), it is clear that \( (m + n)^{th} \) term of the AP is zero. Hence proved.

Question. The \( 19^{th} \) term of an AP is equal to three times its sixth term. If its \( 9^{th} \) term is 19, find the AP.
Answer: 3, 5, 7, 9, \dots

Question. The sum of the \( 4^{th} \) and \( 8^{th} \) terms of an AP is 24 and the sum of the \( 6^{th} \) and \( 10^{th} \) terms is 44. Find the first three terms of the AP.
Answer: -13, -8, -3

Question. The eighth term of an AP is half its second term and the eleventh term exceeds one-third of its fourth term by 1. Find the \( 15^{th} \) term.
Answer: 3

Question. If 4 times the \( 4^{th} \) term of an AP is equal to 18 times the \( 18^{th} \) term, then find the \( 22^{nd} \) term.
Answer: 0