CBSE Class 10 Arithmetic Progressions Sure Shot Questions Set 16

Evaluation and Analysis Based Questions

Question. If the \( p^{th} \) term of an A.P. is \( \frac{1}{q} \) and \( q^{th} \) term is \( \frac{1}{p} \), prove that the sum of the first \( pq \) term, is \( \frac{1}{2} ( pq + 1) \).
Answer: Let \( a, d \) be the first term and the common difference, respectively. It is given that
\( T_p = \frac{1}{q} \)
\( \Rightarrow a + (p - 1)d = \frac{1}{q} \) ...(i)
and \( T_q = \frac{1}{p} \)
\( \Rightarrow a + (q - 1)d = \frac{1}{p} \) ...(ii)
Subtracting equation (ii) from equation (i), we get
\( (q - 1 - p + 1) d = \frac{1}{p} - \frac{1}{q} \)
\( \Rightarrow (q - p) d = \frac{q - p}{pq} \)
\( \Rightarrow d = \frac{1}{pq} \) ...(iii)
Put the value of \( d \) in equation (i), we get
\( a + \frac{p - 1}{pq} = \frac{1}{q} \)
\( \Rightarrow a = \frac{1}{q} - \frac{p - 1}{pq} = \frac{p - p + 1}{pq} = \frac{1}{pq} \)
\( \therefore a = \frac{1}{pq} \) ...(iv)
Now, we know \( S_n = \frac{n}{2} [2a + (n - 1)d] \)
\( \therefore S_{pq} = \frac{pq}{2} \left[ \frac{2}{pq} + (pq - 1)\frac{1}{pq} \right] \)
\( = \frac{pq}{2} \left[ \frac{2 + pq - 1}{pq} \right] \)
\( = \frac{1}{2}(pq + 1) \)
Hence proved.

Question. Find the sum of the first 15 multiples of 8.
Answer: The first 15 multiples of 8 are 8, 16, 24, ..... 120
Clearly, these numbers are in A.P. with first term, \( a = 8 \) and common difference, \( d = 16 - 8 = 8 \)
Thus, \( S_{15} = \frac{15}{2} [2 \times 8 + (15 - 1) \times 8] \)
\( = \frac{15}{2} [16 + 14 \times 8] \)
\( = \frac{15}{2} [16 + 112] \)
\( = \frac{15}{2} \times 128 \)
\( = 15 \times 64 = 960 \).

Question. Find the sum of all two-digit natural numbers which when divided by 3 yield 1 as remainder.
Answer: Two-digit natural numbers which when divided by 3 yield 1 as remainder are : 10, 13, 16, 19, .......... 97,
Clearly these numbers are in A.P. with \( a = 10, d = 3, a_n = 97 \)
Now, \( a_n = 97 \)
\( \Rightarrow a + (n - 1)d = 97 \)
\( \Rightarrow 10 + (n - 1)3 = 97 \)
\( \Rightarrow (n - 1) = \frac{87}{3} = 29 \)
\( \Rightarrow n = 30 \)
So, \( S_{30} = \frac{30}{2} [2 \times 10 + (30 - 1) \times 3] \)
\( = 15 [20 + 29 \times 3] \)
\( = 15 [20 + 87] = 15 \times 107 = 1605 \).

Question. If the \( m^{th} \) term of an A.P. is \( \frac{1}{n} \) and \( n^{th} \) term is \( \frac{1}{m} \), then show that its \( (mn)^{th} \) term is 1.
Answer: Let \( a \) and \( d \) be the first term and the common difference, respectively of the given A.P. Then,
\( a_m = a + (m - 1)d \Rightarrow a + (m - 1)d = \frac{1}{n} \) ...(i)
and \( a_n = a + (n - 1)d \Rightarrow a + (n - 1)d = \frac{1}{m} \) ...(ii)
Subtracting equation (ii) from equation (i), we get
\( (m - n) d = \frac{1}{n} - \frac{1}{m} \Rightarrow (m - n) d = \frac{m - n}{mn} \)
\( \Rightarrow d = \frac{1}{mn} \)
Putting \( d = \frac{1}{mn} \) in equation (i), we get
\( a + (m - 1) \frac{1}{mn} = \frac{1}{n} \Rightarrow a + \frac{m}{mn} - \frac{1}{mn} = \frac{1}{n} \)
\( \Rightarrow a - \frac{1}{mn} = 0 \Rightarrow a = \frac{1}{mn} \)
\( \therefore (mn)^{th} \) term \( = a + (mn - 1)d \)
\( = \frac{1}{mn} + (mn - 1) \frac{1}{mn} \)
\( = \frac{1 + mn - 1}{mn} = 1 \)
\( \therefore (mn)^{th} \) term = 1. Hence Proved.

Question. If the sum of \( m \) terms of A.P. is the same as the sum of its \( n \) terms, show that the sum of its \( (m + n) \) terms is zero.
Answer: Let \( a \) be the first term and \( d \) be the common difference of the given A.P. [Given]
Then, \( S_m = S_n \)
\( \Rightarrow \frac{m}{2} \{2a + (m - 1)d\} = \frac{n}{2} \{2a + (n - 1)d\} \)
\( \Rightarrow 2am + m(m - 1) d = 2an + n (n - 1)d \)
\( \Rightarrow 2a(m - n) + \{m (m - 1) - n (n - 1)\} d = 0 \)
\( \Rightarrow 2a (m - n) + \{(m^2 - n^2) - (m - n)\}d = 0 \)
\( \Rightarrow (m - n) \{2a + (m + n - 1) d\} = 0 \)
\( \Rightarrow 2a + (m + n - 1) d = 0 \) ...(i)
Now, \( S_{m+n} = \frac{m + n}{2} \{2a + (m + n - 1)d\} \)
\( \Rightarrow S_{m+n} = \frac{m + n}{2} \times 0 = 0 \) [Using (i)]
\( \therefore \) Sum of its \( (m + n) \) term is 0. Hence Proved.

Question. The ratio of the sum of \( m \) and \( n \) terms of an A.P. is \( m^2 : n^2 \). Show that the ratio of the \( m^{th} \) and \( n^{th} \) terms is \( (2m - 1) : (2n - 1) \).
Answer: Let \( a \) be the first term and \( d \) the common difference of the given A.P. Then,
\( S_m = \frac{m}{2} \{2a + (m - 1)d\} \) and \( S_n = \frac{n}{2} \{2a + (n - 1)d\} \)
Given, \( \frac{S_m}{S_n} = \frac{m^2}{n^2} \)
\( \Rightarrow \frac{\frac{m}{2} \{2a + (m - 1)d\}}{\frac{n}{2} \{2a + (n - 1)d\}} = \frac{m^2}{n^2} \)
\( \Rightarrow \frac{2a + (m - 1)d}{2a + (n - 1)d} = \frac{m}{n} \)
\( \Rightarrow \{2a + (m - 1)d\} n = \{2a + (n - 1) d\} m \)
\( \Rightarrow 2a (n - m) = d \{(n - 1) m - (m - 1) n\} \)
\( \Rightarrow 2a (n - m) = d (n - m) \)
\( \Rightarrow d = 2a \)
\( \therefore \frac{a_m}{a_n} = \frac{a + (m - 1)d}{a + (n - 1)d} = \frac{a + (m - 1)2a}{a + (n - 1)2a} \)
\( = \frac{2m - 1}{2n - 1} \). Hence Proved.

Question. The sum of \( n, 2n, 3n \) terms of an A.P. are \( S_1, S_2 \) and \( S_3 \) respectively. Prove that \( S_3 = 3(S_2 - S_1) \).
Answer: Let \( a \) be the first term and \( d \) be the common difference of the A.P.
\( \therefore S_1 = \frac{n}{2}[2a + (n - 1)d] \) ...(i)
\( S_2 = \frac{2n}{2}[2a + (2n - 1)d] \) ...(ii)
\( S_3 = \frac{3n}{2}[2a + (3n - 1)d] \) ...(iii)
Now, \( S_2 - S_1 = \frac{2n}{2}[2a + (2n - 1)d] - \frac{n}{2}[2a + (n - 1)d] \)
\( = \frac{n}{2} [2\{2a + (2n - 1)d\} - \{2a + (n - 1) d\}] \)
\( = \frac{n}{2} [4a + 4nd - 2d - 2a - nd + d] \)
\( = \frac{n}{2} [2a + 3nd - d] \)
\( = \frac{n}{2} [2a + (3n - 1)d] \)
\( \therefore 3(S_2 - S_1) = \frac{3n}{2} [2a + (3n - 1)d] = S_3 \) [Using (iii)]
\( \therefore S_3 = 3(S_2 - S_1) \). Hence Proved.

Question. If \( a^2, b^2, c^2 \) are in A.P., prove that \( \frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b} \) are in A.P.
Answer: Given, \( a^2, b^2, c^2 \) are in A.P.
\( \therefore b^2 - a^2 = c^2 - b^2 \) ...(i)
Now, \( \frac{b}{c+a} - \frac{a}{b+c} = \frac{b^2 + bc - ac - a^2}{(a+c)(b+c)} \)
\( = \frac{(b^2 - a^2) + c(b - a)}{(a+c)(b+c)} \)
\( = \frac{(b - a)(b + a + c)}{(a+b)(b+c)} \) ...(ii)
Also, \( \frac{c}{a+b} - \frac{b}{c+a} = \frac{c^2 + ac - ba - b^2}{(a+b)(c+a)} \)
\( = \frac{c^2 - b^2 + a(c - b)}{(a+b)(c+a)} \)
\( = \frac{(c - b)(c + b + a)}{(a+b)(a+c)} \)
\( = \frac{(c^2 - b^2)(c + b + a)}{(a+b)(b+c)(c+a)} \) [Multiplying numerator and denominator by \( b + c \)]
\( = \frac{(b^2 - a^2)(a + b + c)}{(a+b)(b+c)(c+a)} \) [Using (i)]
\( = \frac{(b - a)(a + b + c)}{(b+c)(c+a)} \) ...(iii)
From equations (ii) and (iii), we have
\( \frac{b}{c+a} - \frac{a}{b+c} = \frac{c}{a+b} - \frac{b}{c+a} \)
\( i.e., \frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b} \) are in A.P. Hence Proved.

Question. The sum of the third and seventh terms of an A.P. is 6 and their product is 8. Find the sum of first sixteen terms of the A.P.
Answer: Let \( a \) be the first term and \( d \) be the common difference of the A.P.
We have, \( a_3 + a_7 = 6 \) ...(i) and \( a_3 a_7 = 8 \) ...(ii)
From equation (i), we have
\( \Rightarrow (a + 2d) + (a + 6d) = 6 \)
\( \Rightarrow 2a + 8d = 6 \)
\( \Rightarrow a + 4d = 3 \)
\( \Rightarrow a = 3 - 4d \) ...(iii)
From equation (ii), we have
\( (a + 2d) (a + 6d) = 8 \)
\( \Rightarrow (3 - 4d + 2d) (3 - 4d + 6d) = 8 \) [Using (iii)]
\( \Rightarrow (3 - 2d ) (3 + 2d ) = 8 \)
\( \Rightarrow 9 - 4d^2 = 8 \Rightarrow 4d^2 = 1 \Rightarrow d^2 = \frac{1}{4} \Rightarrow d = \pm \frac{1}{2} \)
Case I : When \( d = \frac{1}{2} \)
Putting \( d = \frac{1}{2} \) in equation (iii), we get
\( a = 3 - 4 \times \frac{1}{2} = 3 - 2 = 1 \)
Now, we know \( S_n = \frac{n}{2} [2a + (n - 1)d] \)
\( S_{16} = \frac{16}{2} [2 \times 1 + (16 - 1)\frac{1}{2}] \)
\( = 8 [2 + 15(\frac{1}{2})] \)
\( = 8 [\frac{4 + 15}{2}] = \frac{16}{2} \times \frac{19}{2} = 4 \times 19 = 76 \)
Case II : When \( d = -\frac{1}{2} \)
Putting \( d = -\frac{1}{2} \) in equation (iii), we get
\( a = 3 - 4(\frac{-1}{2}) = 3 + 2 = 5 \)
\( \therefore S_{16} = \frac{16}{2} [2 \times 5 + (16 - 1)(\frac{-1}{2})] \)
\( = 8 [10 + 15(\frac{-1}{2})] \)
\( = 8 [\frac{20 - 15}{2}] = \frac{16}{2} \times \frac{5}{2} = 4 \times 5 = 20 \).

Question. If in an A.P., the sum of \( m \) terms is equal to \( n \) and the sum of \( n \) terms is equal to \( m \), then prove that the sum of \( (m + n) \) terms is \( -(m + n) \).
Answer: Let \( a \) be the first term and \( d \) be the common difference of the given A.P. Then,
\( S_m = n \) [Given]
\( \Rightarrow \frac{m}{2} \{2a + (m - 1)d\} = n \)
\( \Rightarrow 2am + m (m - 1)d = 2n \) ...(i)
and \( S_n = m \) [Given]
\( \Rightarrow \frac{n}{2} \{2a + (n - 1)d\} = m \)
\( \Rightarrow 2an + n(n - 1)d = 2m \) ...(ii)
Subtracting equation (ii) from equation (i), we get
\( 2a (m - n) + \{m (m - 1) - n (n - 1)\} d = 2n - 2m \)
\( \Rightarrow 2a (m - n) + \{(m^2 - n^2) - (m - n)\} d = - 2(m - n) \)
\( \Rightarrow 2a + (m + n - 1) d = - 2 \) ...(iii) [On dividing both sides by \( (m - n) \)]
Now, \( S_{m+n} = \frac{m + n}{2} [2a + (m + n - 1)d] \)
\( \Rightarrow S_{m+n} = \frac{m + n}{2} (-2) \) [Using (iii)]
\( \Rightarrow S_{m+n} = - (m + n) \). Hence Proved.

Question. The sum of the first \( p, q, r \) terms of an A.P. are \( a, b, c \) respectively. Show that \( \frac{a}{p}(q - r) + \frac{b}{q}(r - p) + \frac{c}{r}(p - q) = 0 \).
Answer: Let \( A \) be the first term and \( D \) be the common difference of the given A.P. Then,
\( a = \) Sum of \( p \) terms
\( \Rightarrow a = \frac{p}{2} \{2A + (p - 1) D\} \)
\( \Rightarrow \frac{2a}{p} = \{2A + (p - 1) D\} \) ...(i)
\( b = \) Sum of \( q \) terms
\( \Rightarrow b = \frac{q}{2} \{2A + (q - 1) D\} \) ...(ii)
\( \Rightarrow \frac{2b}{q} = \{2A + (q - 1) D\} \)
and \( c = \) Sum of \( r \) terms
\( \Rightarrow c = \frac{r}{2} \{2A + (r - 1) D\} \) ...(iii)
\( \Rightarrow \frac{2c}{r} = \{2A + (r - 1) D\} \)
Multiplying equations (i), (ii) and (iii) by \( (q - r) \), \( (r - p) \) and \( (p - q) \) respectively and adding, we get
\( \frac{2a}{p}(q - r) + \frac{2b}{q}(r - p) + \frac{2c}{r}(p - q) = 2A[q - r + r - p + p - q] + D[(p - 1) (q - r) + (q - 1) (r - p) + (r - 1) (p - q)] \)
\( \Rightarrow \frac{a}{p}(q - r) + \frac{b}{q}(r - p) + \frac{c}{r}(p - q) = D[pq - pr - q + r + qr - pq - r + p + pr - rq - p + q] \)
\( \Rightarrow \frac{a}{p}(q - r) + \frac{b}{q}(n - p) + \frac{c}{r}(p - q) = 0 \). Hence Proved.

Question. If there are \( (2n + 1) \) terms in A.P., then prove that the ratio of the sum of odd terms and the sum of even terms is \( (n + 1) : n \).
Answer: Let \( a \) and \( d \) be the first term and common difference respectively of the given A.P.
Let \( a_k \) denote the \( k^{th} \) term of the given A.P. Then, \( a_k = a + (k - 1) d \)
Let \( S_1 = \) Sum of odd terms
\( \Rightarrow S_1 = a_1 + a_3 + a_5 +.....+ a_{2n + 1} \)
\( \Rightarrow S_1 = \frac{n + 1}{2} \{a_1 + a_{2n + 1}\} \)
\( \Rightarrow S_1 = \frac{n + 1}{2} \{a + a + (2n + 1 - 1) d\} \) [\( \because a_{2n + 1} = a + (2n + 1 - 1) d \)]
\( \Rightarrow S_1 = (n + 1) (a + nd) \)
and, \( S_2 = \) Sum of even terms
\( \Rightarrow S_2 = a_2 + a_4 + a_6 +.....+ a_{2n} \)
\( \Rightarrow S_2 = \frac{n}{2} [a_2 + a_{2n}] \)
\( \Rightarrow S_2 = \frac{n}{2} [(a + d ) + \{a + (2n - 1) d\}] \) [\( \because a_{2n} = a + (2n - 1) d \)]
\( \Rightarrow S_2 = n (a + nd ) \)
\( \therefore S_1 : S_2 = (n + 1) (a + nd) : n(a + nd) = (n + 1) : n \). Hence Proved.

Question. Find the sum of the integers lying between 1 and 100 (both inclusive) and divisible by 3, 5 or 7.
Answer: Integers divisible by 3 from 1 to 100 are 3, 6, 9,.....99, \( i.e., \) total 33 in number
Integers divisible by 5 are, 5, 10, 15,........100 \( i.e., \) total 20 in number
Integers divisible by 7 are 7, 14,..., 98, \( i.e., \) total 14 in number
Integers divisible by both 3 and 5 are 15, 30........90, \( i.e., \) total 6 in number
Integers divisible by both 3 and 7 are 21, 42, 63, and 84. \( i.e., \) total 4 in number
and Integers divisible by both 5 and 7 are 35, 70 \( i.e., \) total 2 in number
So, sum of numbers divisible by 3, 5 or 7 is
\( = \frac{33}{2} (3 + 99) + \frac{20}{2} (5 + 100) + \frac{14}{2} (7 + 98) - \frac{6}{2} (15 + 90) - \frac{4}{2} (21 + 84) - \frac{2}{2} (35 + 70) \)
\( = \frac{33}{2} \times 102 + \frac{20}{2} \times 105 + \frac{14}{2} \times 105 - \frac{6}{2} \times 105 - \frac{4}{2} \times 105 - \frac{2}{2} \times 105 \)
\( = 33 \times 51 + 105 ( 10 + 7 - 3 - 2 - 1) \)
\( = 1683 + 105 \times 11 \)
\( = 1683 + 1155 \)
\( = 2838 \).

Question. The digits of a positive integer having three digits are in A.P. and their sum is 15. If the number obtained by reversing the digits is 594 less than the original number, then find the number.
Answer: Let \( a \) and \( d \) be the first term and the common difference of the A.P. Then, let \( (a - d), a, (a + d) \) be the digits of three-digit number.
Then, \( a - d + a + a + d = 15 \) [Given]
\( \Rightarrow 3a = 15 \)
\( \Rightarrow a = 5 \)...(i)
According to the question,
Original number \( = 100(a - d) + 10a + (a + d) \)
\( = 111a - 99d \)
\( = 111 \times 5 - 99d \) [Using (i)]
\( = 555 - 99d \)
Number formed by reversing the digits \( = 100(a + d) + 10a + (a - d) \)
\( = 111a + 99d \) [Using (i)]
\( = 111 \times 5 + 99d \)
\( = 555 + 99d \)
\( \therefore (555 - 99d) - (555 + 99d) = 594 \) [Given]
\( \Rightarrow -198d = 594 \)
\( \Rightarrow d = -3 \)
Thus, the digit in the unit’s place is \( 5 - 3 = 2 \), in the ten’s place is 5 and in the hundred’s place is \( 5 + 3 = 8 \). Hence the number is 852.

Assertion and Reasoning Based Questions

Question. Assertion: If the sum of \( n \) terms of a series is \( 2n^2 + 3n + 1 \) then series is in A.P. with common difference 4.
Reason: If sum of \( n \) terms of an A.P is quadratic expression, then common difference is twice of the coefficient of quadratic term.
(a) Both the Assertion and the Reason are correct and Reason is the correct explanation of the Assertion.
(b) Both the Assertion and the Reason are correct but Reason is not the correct explanation of the Assertion.
(c) Assertion is true but Reason is false.
(d) Both Assertion and Reason are false.
Answer: (A) Both the Assertion and the Reason are correct and Reason is the correct explanation of the Assertion.
Explanation :
As per the reason statement,
Sum of \( n \) terms of A.P. is \( S_n = \frac{n}{2} [2A + (n - 1)D] \)
Where A = first term, D = common difference. Hence, the sum of \( n \) terms of an A.P. is always in the form of quadratic expression.
Hence, it is proved that reason is true.
As per the assertion given,
\( S_n = 2n^2 + 3n + 1 \)
\( \therefore t_n = S_n - S_{n-1} \)
\( = 2[n^2 - (n - 1)^2] + 3[n - n + 1] \)
\( = 2[2n - 1] + 3 = 4n + 1 \)
\( \therefore D = t_{n+1} - t_n = 4 \)
So, both assertion and reason are correct and reason in correct explanation of assertion.

Question. Assertion: If sum of \( n \) terms of two arithmetic progressions are in the ratio \( (3n + 8): (7n + 15) \) then ratio of their \( n^{th} \) term is 3:16
Reason: If \( S_n \) is quadratic expression, then \( t_n = S_n - S_{n-1} \).
(a) Both the Assertion and the Reason are correct and Reason is the correct explanation of the Assertion.
(b) Both the Assertion and the Reason are correct but Reason is not the correct explanation of the Assertion.
(c) Assertion is true but Reason is false.
(d) Both Assertion and Reason are false.
Answer: (C) Assertion is false but Reason is true.
Explanation :
\( \frac{S_n}{S'_n} = \frac{3n + 8}{7n + 15} = \frac{n(3n+8)}{n(7n+15)} = \frac{3n^2 + 8n}{7n^2 + 15n} \)
\( \Rightarrow \frac{t_n}{t'_n} = \frac{S_n - S_{n-1}}{S'_n - S'_{n-1}} = \frac{(3n^2 + 8n) - 3(n-1)^2 - 8(n-1)}{7n^2 + 15n - 7(n-1)^2 - 15(n-1)} \)
\( \Rightarrow \frac{t_n}{t'_n} = \frac{6n + 5}{14n + 8} \)
\( \Rightarrow t_n : t'_n = 6n + 5 : 14n + 8 \)
So, assertion is proved to be false and reason is true.

Question. Assertion: The sum of the series with the \( n^{th} \) term, \( t_n = (9 - 5n) \) is – 465.
Reason: Given series is in A.P. and sum of \( n \) terms of an A.P. is \( S_n = \frac{n}{2} [2a + (n - 1)d] \)
(a) Both the Assertion and the Reason are correct and Reason is the correct explanation of the Assertion.
(b) Both the Assertion and the Reason are correct but Reason is not the correct explanation of the Assertion.
(c) Assertion is true but Reason is false.
(d) Both Assertion and Reason are false.
Answer: (B) Both the Assertion and the Reason are correct but Reason is not the correct explanation of the Assertion.
Explanation :
When \( n = 1 \), then \( t_1 = 4 \)
When \( n = 15 \) then \( t_{15} = - 66 \)
Sum
\( \frac{n}{2}(a+l) = \frac{15}{2}(4-66) \)
\( = \frac{15(- 62)}{2} \)
\( = - 465 \)
So, both assertion and reason are correct and reason is the correct explanation of the assertion.

Case Based Questions

The production of TV sets in a factory increases uniformly by a fixed number every year. It produced 16000 sets in 6th year and 22600 in 9th year.

Question. What is the production during first year ?
(a) 5000
(b) 2200
(c) 10000
(d) None of the options
Answer: (a) 5000
Explanation :
Let the production during first year be \( a \) and let \( d \) be the increase in productoin every year. Then,
\( a_6 = 16000 \)
\( \Rightarrow a + 5d = 16000 \)...(i)
and \( a_9 = 22600 \)
\( \Rightarrow a + 8d = 22600 \)...(ii)
On subracting (i) from (ii), we get
\( 3d = 6600 \)
\( \Rightarrow d = 2200 \).
Putting \( d = 2200 \) in (i), we get \( a + 5 \times 2200 = 16000 \)
\( \Rightarrow a + 11000 = 16000 \)
\( \Rightarrow a = 16000 - 11000 = 5000 \).
Thus, \( a = 5000 \) and \( d = 2200 \).

Question. Find the production during 8th year.
(a) 7200
(b) 22000
(c) 20400
(d) None of the options
Answer: (c) Rs. 20400.
Explanation :
Production during 8th year is given by \( a_8 \)
\( = (a + 7d) \)
\( = (5000 + 7(2200)) \)
\( = (5000 + 15400) \)
\( = 20400 \).

Question. Find the production during first 3 years is :
(a) Rs. 21600
(b) Rs. 22000
(c) Rs. 20400
(d) None of the options
Answer: (a) Rs. 21600.
Explanation :
\( a_2 = (a + d) = (5000 + 2200) = 7200 \).
\( a_3 = (a_2 + d) = 7200 + 2200 = 9400 \).
\( \therefore \) Production during first 3 years
\( = 5000 + 7200 + 9400 \)
\( = 21600 \)

Question. In which year, the production is Rs. 29,200.
(a) 10
(b) 11
(c) 12
(d) 13
Answer: (c) 12.
Explanation :
\( a_n = 5000 + (n - 1)2200 = 29200 \)
\( (n - 1)2200 = 29200 - 5000 = 24200 \)
\( \Rightarrow n - 1 = 11 \)
\( \Rightarrow n = 12 \)

Question. Find the difference of the production during 7th year and 4th year is.
(a) Rs. 5000
(b) Rs. 2200
(c) Rs. 10000
(d) None of the options
Answer: (d) None of the options.
Explanation :
\( a_4 = (a + 3d) = (5000 + 3(2200)) = 5000 + 6600 = 11600 \).
\( a_7 = (a_6 + d) = 16000 + 2200 = 182000 \).
Difference \( = 18200 - 11600 = 6600 \)

Amit was playing a number card game. In the game, some number cards (having both +ve or –ve numbers) are arranged in a row such that they are following an arithmetic progression. On his first turn, Amit picks up 6th and 14th card and finds their sum to be – 76. On the second turn he picks up 8th and 16th card and finds their sum to be – 96.

Question. What is the difference between the numbers on any two consecutive cards?
(a) 7
(b) – 5
(c) 11
(d) – 3
Answer: (b) – 5.
Explanation :
Let the numbers on the cards be \( a, a + d, a + 2d, \dots \)
According to questions, we have
\( (a + 5d) + (a + 13d) \)
\( \Rightarrow 2a + 18d = - 76 \Rightarrow a + 9d = - 38 \)...(i)

Question. The number on first card is :
(a) 12
(b) 3
(c) 5
(d) 7
Answer: (d) 7
Explanation :
And \( (a + 7d) + (a + 15d) = - 96 \)
\( \Rightarrow 2a + 22d = - 96 \Rightarrow a + 11d = - 48 \)...(ii)
From (i) and (ii), we get
\( 2d = - 10 \Rightarrow d = - 5 \)
From (i), \( a + 9(- 5) = - 38 \Rightarrow a = 7 \)
Number on first card \( = a = 7 \)

Question. What is the number on the \( 19^{th} \) card?
(a) –88
(b) –83
(c) –92
(d) –102
Answer: (b) – 83
Explanation :
Number on \( 19^{th} \) card \( = a + 18d \)
\( = 7 + 18(-5) = - 83 \)

Question. What is the number on the 23rd card?
(a) –103
(b) –122
(c) –108
(d) –117
Answer: (a) – 103
Explanation :
Number on 23rd card \( = a + 22d \)
\( = 7 + 22(-5) = - 103 \)

Question. What is the sum of \( 9^{th} \) and \( 15^{th} \) card ?
(a) –129
(b) –122
(c) –180
(d) –171
Answer: (a) – 129
Explanation :
Number on \( 9^{th} \) card \( = a + 8d \)
\( = 7 + 8 \times - 5 \)
\( = 7 - 40 = - 33 \)
Number on \( 15^{th} \) card \( = a + 14d \)
\( = 7 + 14 \times - 5 \)
\( = 7 - 70 = - 63 \)
Sum of \( 9^{th} \) & \( 15^{th} \) card \( = - 33 + (- 63) \)
\( = - 129 \)

Treasure Hunt Games. While playing a treasure hunt game, some clues (numbers) are hidden in various spots collectively forms an AP. If the number of the nth spot is \( 20 + 4n \), then answer the following questions to help the player in spotting the clues.

Question. Which number is on the first spot ?
(a) 20
(b) 24
(c) 16
(d) 28
Answer: (b) 24.
Explanation :
Number of \( n^{th} \) spot \( = 20 + 4n \)
\( i.e., t_n = 20 + 4n \)
Then, \( t_1 = 20 + 4 \times 1 = 24 \)

Question. Which number is on the \( (n - 2)^{th} \) spot ?
(a) \( 16 + 4n \)
(b) \( 24 + 4n \)
(c) \( 12 + 4n \)
(d) \( 28 + 4n \)
Answer: (c) \( 12 + 4n \).
Explanation :
Number on \( (n - 2)^{th} \) spot \( = t_{n - 2} = 20 + 4(n - 2) \)
\( = 20 + 4n - 8 = 12 + 4n \)

Question. Which number is on the \( 34^{th} \) spot ?
(a) 156
(b) 116
(c) 120
(d) 160
Answer: (a) 156
Explanation :
Number on \( 34^{th} \) spot \( = t_{34} \)
\( = 20 + 4(34) = 156 \)

Question. Which spot is numbered as 116 ?
(a) \( 5^{th} \)
(b) \( 8^{th} \)
(c) \( 9^{th} \)
(d) \( 24^{th} \)
Answer: (d) \( 24^{th} \).
Explanation :
Let \( n^{th} \) spot be numbered as 116.
\( \therefore t_n = 116 \)
\( \Rightarrow 20 + 4n = 116 \)
\( \Rightarrow 4n = 96 \Rightarrow n = 24 \)

A sequence is an ordered list of numbers. A sequence of numbers such that difference between the consecutive terms is constant is said to be an arithmetic progression (A.P.). On the basis of above information answer the following.

Question. Which of the following sequence is an A.P. ?
(a) 10, 24, 39, 52, ...
(b) 11, 24, 39, 52, ...
(c) 10, 24, 38, 52, ...
(d) 10, 38, 52, 66, ...
Answer: (c) 10, 24, 38, 52, ...

Question. If \( x, y \) and \( z \) are in A.P., then ?
(a) \( x + z = y \)
(b) \( x - z = y \)
(c) \( x + z = 2y \)
(d) None of the options
Answer: (c) \( x + z = 2y \).

Question. If \( a_1, a_2, a_3, \dots, a_n \) in A.P. then which of the following is true ?
(a) \( a_1 + k, a_2 + k, a_3 + k, \dots, a_n + k \) are in A.P., where \( k \) is a constant.
(b) \( k - a_1, k - a_2, k - a_3, \dots, k - a_n \) are in A.P., where \( k \) is a constant.
(c) \( ka_1, ka_2, ka_3, \dots, ka_n \) are in A.P., where \( k \) is a constant.
(d) All of the options
Answer: (d) All of the options.

Question. If the \( n^{th} \) term (\( n > 1 \)) of an A.P. is smaller than the first term, then nature of its common difference (\( d \)) is :
(a) \( d > 0 \)
(b) \( d < 0 \)
(c) \( d = 0 \)
(d) Can’t be determined
Answer: (b) \( d < 0 \).

Question. Which of the following is incorrect about A.P. ?
(a) All the terms of constant A.P. are same.
(b) Some terms of an A.P. can be negative.
(c) All the terms of an A.P. can never be negative.
(d) None of the options.
Answer: (c) All the terms of an A.P. can never be negative.

India is competitive manufacturing location due to the low cost of manpower and strong technical and engineering capabilities contributing to higher quality production runs. The production of TV sets in a factory increases uniformly by a fixed number every year. It produced 16000 sets in 6th year and 22600 in 9th year.

Question. Find the production for first year.
Answer: \( a_6 = 16000 \)
and \( a_9 = 22600 \)
Since, \( a_n = a + (n - 1)d \)
\( \therefore a_6 = a + 5d = 16000 \)...(i)
and \( a_9 = a + 8d = 22600 \)...(ii)
Solving equations (i) and (ii), we get
\( d = 2200 \)
and \( a = 5000 \)

Question. Find the production during \( 8^{th} \) year.
Answer: \( a_8 = a + 7d \)
Production during \( 8^{th} \) year is \( (a + 7d) \)
\( = 5000 + 7(2200) = 20400 \)

Question. Find the production during first 3 years.
Answer: Production during first 3 years
\( = (5000) + (5000 + 2200) + (5000 + 2 \times 2200) \)
\( = 21600 \)

Question. In which year, the production is ₹ 29,200.
Answer: Given, \( a_n = 29200 \)
\( \Rightarrow a + (n - 1)d = 29200 \)
\( \Rightarrow 5000 + (n - 1)2200 = 29200 \)
\( \Rightarrow (n - 1)2200 = 24200 \)
\( \Rightarrow n = 12 \)

Question. Find the difference of the production during \( 7^{th} \) year and \( 4^{th} \) year.
Answer: \( a_7 - a_4 = (a + 6d) - (a + 3d) \)
\( = a + 6d - 3d) \)
\( = 3d \)
\( = 3 \times 2200 = 6600 \)

Your friend Veer wants to participate in a 200 m race. He can currently run that distance in 51 seconds and with each day of practice it takes him 2 seconds less. He wants to do in 31 seconds. He win it

Question. Which of the following terms are in A.P. for the given situation.
(a) 51, 53, 55 ...
(b) 51, 49, 47 ....
(c) – 51, – 53, – 55 ...
(d) 51, 55, 59 ...
Answer: (b) 51, 49, 47 ...
Explanation :
A.P. \( = 51, (51 - 2), (51 - 2 - 2) \dots (31) \)
A.P. \( = 51, 49, 47, \dots 31 \)

Question. What is the minimum number of days he needs to practice till his goal is achieved.
(a) 10
(b) 12
(c) 11
(d) 9
Answer: (c) 11
Explanation :
\( a = 51, d = - 2 \)
Here, \( a_n = 31 \)
\( \Rightarrow a_n = a + (n - 1)d \)
\( 31 = 51 + (n - 1)(- 2) \)
\( n = 11 \) days
So, he will achieve his target in 11 days

Question. Which of the following term is not in the A.P. of the above given situation.
(a) 41
(b) 30
(c) 37
(d) 39
Answer: (b) 30
Explanation :
As, 31 seconds is the last time he need to achieve.

Question. If \( n^{th} \) term of an A.P. is given by \( a_n = 2n + 3 \), then common difference of an AP.
(a) 2
(b) 3
(c) 5
(d) 1
Answer: (a) 2
Explanation :
\( a_n = 2n + 3 \)
\( a_1 = 2 \times 1 + 3 = 5 \)
\( a_2 = 2 \times 2 + 3 = 7 \)
\( a_3 = 2 \times 3 + 3 = 9 \)
\( a_4 = 2 \times 4 + 3 = 11 \)
A.P. = 5, 7, 9, 11
\( d = 7 - 5 = 2 \)

Question. The value of \( x \), for which \( 2x, x + 10, 3x + 2 \) are three consecutive terms of an AP
(a) 6
(b) – 6
(c) 18
(d) – 18
Answer: (a) 6
Explanation :
\( a = 2x, b = x + 10, c = 3x + 2 \)