CBSE Class 10 Arithmetic Progressions Sure Shot Questions Set 15

Question. Find the common difference of the A.P. \( \frac{1}{3q}, \frac{1-6q}{3q}, \frac{1-12q}{3q} \dots \)
Answer: Common difference,
\( d = a_2 - a_1 = \frac{1 - 6q}{3q} - \frac{1}{3q} \)
\( = \frac{1 - 6q - 1}{3q} = \frac{- 6q}{3q} = - 2 \)

Question. The angles of a triangle are in A.P., the least being half the greatest. Find the angles.
Answer: Let the angles be \( a - d, a, a + d \); \( a > 0, d > 0 \)
\( \therefore \) Sum of angles = \( 180^{\circ} \)
\( \therefore a - d + a + a + d = 180^{\circ} \)
\( \Rightarrow 3a = 180^{\circ} \)
\( \therefore a = 60^{\circ} \)...(i)
By the given condition
\( a - d = \frac{a+d}{2} \)
\( \Rightarrow 2a - 2d = a + d \)
\( \Rightarrow 2a - a = d + 2d \)
\( \Rightarrow a = 3d \)
\( \Rightarrow d = \frac{a}{3} = \frac{60}{3} = 20^{\circ} \)[From (i)]
\( \therefore \) Angles are \( 60^{\circ} - 20^{\circ}, 60^{\circ}, 60^{\circ} + 20^{\circ} \) \( i.e., 40^{\circ}, 60^{\circ}, 80^{\circ} \).

Question. Which term of the progression 4, 9, 14, 19, .... is 109?
Answer: Here, \( d = 9 - 4 = 14 - 9 = 19 - 14 = 5 \)
\( \therefore \) Difference between consecutive terms is constant.
Hence it is an A.P.
Given : First term, \( a = 4, d = 5, a_n = 109 \)
Since, \( a_n = a + (n - 1)d \)
\( \therefore 109 = 4 + (n - 1)5 \)
\( \Rightarrow 105 = 5(n - 1) \)
\( \Rightarrow n - 1 = \frac{105}{5} = 21 \)
\( \Rightarrow n = 21 + 1 = 22 \)
\( \therefore 109 \) is the \( 22^{nd} \) term.

Question. How many natural numbers are there between 200 and 500, which are divisible by 7?
Answer: Natural numbers, divisible by 7 between 200 and 500 is,
203, 210, 217, ..., 497
Here, \( a = 203, d = 210 - 203 = 7, a_n = 497 \)
\( \therefore a + (n - 1)d = a_n \)
\( \Rightarrow 203 + (n - 1)7 = 497 \)
\( \Rightarrow (n - 1)7 = 497 - 203 = 294 \)
\( \Rightarrow n - 1 = \frac{294}{7} = 42 \)
\( \therefore n = 42 + 1 = 43 \)
\( \therefore \) There are 43 natural numbers between 200 and 500 which are divisible by 7.

Question. The sum of the 5th and the 9th terms of an A.P. is 30. If its 25th term is three times its 8th term, find the A.P.
Answer: \( a_5 + a_9 = 30 \)[Given]
\( \therefore a + 4d + a + 8d = 30 \) [\( \because a_n = a + (n - 1)d \)]
\( \Rightarrow 2a + 12d = 30 \)
\( \Rightarrow a + 6d = 15 \)
\( a = 15 - 6d \)...(i)
Now, \( a_{25} = 3(a_8) \)
\( \Rightarrow a + 24d = 3(a + 7d) \)
\( \Rightarrow 15 - 6d + 24d = 3(15 - 6d + 7d) \)
[From (i)]
\( \Rightarrow 15 + 18d = 3(15 + d) \)
\( \Rightarrow 15 + 18d = 45 + 3d \)
\( \Rightarrow 18d - 3d = 45 - 15 \)
\( 15d = 30 \)
\( \therefore d = \frac{30}{15} = 2 \)
From (i),
\( a = 15 - 6(2) = 15 - 12 = 3 \)
\( \therefore \) A.P. is \( a, a + d, a + 2d, a + 3d, \dots \)
\( = 3, 5, 7, 9, \dots \)

Question. The digits of a positive number of three digits are in A.P. and their sum is 15. The number obtained by reversing the digits is 594 less than the original number. Find the number.
Answer: Let hundred‘s place digit = \( (a - d) \)
Let ten‘s place digit = \( a \)
Let unit‘s place digit = \( a + d \)
According to the question,
\( a - d + a + a + d = 15 \)
\( \Rightarrow 3a = 15 \)
\( \Rightarrow a = 5 \)
Original number
\( = 100(a - d) + 10(a) + 1(a + d) \)
\( = 100a - 100d + 10a + a + d \)
\( = 111a - 99d \)
Reversed number
\( = 1(a - d) + 10a + 100(a + d) \)
\( = a - d + 10a + 100a + 100d \)
\( = 111a + 99d \)
Now, original no. – Reversed no. = 594
\( 111a - 99d - (111a + 99d) = 594 \)
\( \Rightarrow -198d = 594 \)
\( \Rightarrow d = \frac{594}{-198} \)
\( = - 3 \)
\( \therefore \) Original number = \( 111a - 99d \)
\( = 111(5) - 99(- 3) \)
\( = 555 + 297 = 852 \)

Question. If \( p^{th} \), \( q^{th} \) and \( r^{th} \) terms of an A.P. are \( a, b, c \) respectively, then show that \( (a - b)r + (b - c)p + (c - a)q = 0 \).
Answer: Let \( A \) be the first term and \( D \) be the common difference of the given A.P.
\( p^{th} \) term = \( A + (p - 1)D = a \)...(i)
\( q^{th} \) term = \( A + (q - 1)D = b \)...(ii)
\( r^{th} \) term = \( A + (r - 1)D = c \)...(iii)
\( L.H.S. = (a - b)r + (b - c)p + (c - a)q \)
\( = [A + (p - 1)D - (A + (q - 1)D)]r \)
\( + [A + (q - 1)D - (A + (r - 1)D)]p \)
\( + [A + (r - 1)D - (A + (p - 1)D)]q \)
\( = [(p - 1 - q + 1)D]r + [(q - 1 - r + 1)D]p + [(r - 1 - p + 1)D]q \)
\( = D[(p - q)r + (q - r)p + (r - p)q] \)
\( = D[pr - qr + qp - rp + rq - pq] \)
\( = D[0] = 0 = R.H.S. \)

Short Answer Type Questions

Question. If the \( 8^{th} \) term of an A.P. is 31 and the \( 15^{th} \) term is 16 more than the \( 11^{th} \) term, find the A.P.*
Answer: Let the first term be \( a \) and common difference be \( d \).
Thus, \( t_8 = a + (8 - 1)d = 31 \)
\( \Rightarrow a + 7d = 31 \)…(i)
and \( t_{15} = 16 + t_{11} \)
\( \Rightarrow a + (15 - 1)d = 16 + a + (11 - 1)d \)
\( \Rightarrow a + 14d = 16 + a + 10d \)
\( \Rightarrow 4d = 16 \)
\( \Rightarrow d = 4 \)
Substituting \( d = 4 \) in (i), we get
\( a + 7(4) = 31 \)
\( \Rightarrow a + 28 = 31 \)
\( \Rightarrow a = 3 \)
Thus, the A.P. is 3, 7, 11, 15 …

Question. Which term of the A.P. 8, 14, 20, … will be 72 more than the \( 41^{st} \) term ?*
Answer: Given, \( a = 8, d = 6 \)
Thus, \( t_{41} = 8 + (41 - 1)6 \)
\( \Rightarrow t_{41} = 8 + 240 \)
\( \Rightarrow t_{41} = 248 \)
Now, \( 72 + 248 = 320 \)
Let the \( n^{th} \) term be 320.
So, \( t_n = 8 + (n - 1)6 = 320 \)
\( \Rightarrow 2 + 6n = 320 \)
\( \Rightarrow 6n = 318 \)
\( \Rightarrow n = 53 \).
Thus, the required term is 53 term.

Question. The sum of the \( 5^{th} \) and the \( 9^{th} \) term of an A.P. is 30. If its \( 25^{th} \) term is three times the \( 8^{th} \) term, find the A.P.*
Answer: Let \( a \) be the first term and \( d \) be the common difference.
\( \therefore T_5 = a + (5 - 1)d \)
\( \Rightarrow T_5 = a + 4d \)
and \( T_9 = a + (9 - 1)d \)
\( \Rightarrow T_9 = a + 8d \)
Now, \( T_5 + T_9 = 30 \) [Given]
\( \Rightarrow a + 4d + a + 8d = 30 \)
\( \Rightarrow 2a + 12d = 30 \)
\( \Rightarrow a + 6d = 15 \)…(i)
Also, \( T_{25} = a + (25 - 1)d \)
\( \Rightarrow T_{25} = a + 24d \)
and \( T_8 = a + (8 - 1)d \)
\( \Rightarrow T_8 = a + 7d \)
Now, \( T_{25} = 3T_8 \)[Given]
\( \Rightarrow a + 24d = 3(a + 7d ) \)
\( \Rightarrow a + 24d = 3a + 21d \)
\( \Rightarrow 2a - 3d = 0 \)
On multiplying the above equation by 2, we get
\( 4a - 6d = 0 \)…(ii)
Adding equations (i) and (ii), we get
\( 5a = 15 \)
\( \Rightarrow a = 3 \)
and \( d = \frac{2(3)}{3} = 2 \)
So the required A.P. is 3, 5, 7, 9, …

Question. The sum of the first three terms of an A.P. is 48. If the product of the first and the second terms exceeds four times the third term by 12, find the A.P.*
Answer: Given, \( S_3 = 48 \)
\( S_3 = \frac{3}{2} \{2a + (3 - 1)d\} \)
\( \Rightarrow 48 = \frac{3}{2} \{2a + 2d\} \)
\( \Rightarrow 48 = 3(a + d) \)
\( \Rightarrow a + d = 16 \)
\( \Rightarrow d = 16 - a \)
Now, \( T_2 = a + (2 - 1) d = a + d \)
\( T_3 = a + (3 - 1) d = a + 2d \)
Now, \( a(a + d) = 4(a + 2d ) + 12 \)[Given]
\( \Rightarrow 16a = 4a + 8d + 12 \)[\( \because a + d = 16 \)]
\( \Rightarrow 12a = 8d + 12 \)
\( \Rightarrow 3a = 2d + 3 \)
\( \Rightarrow 3a = 2(16 - a) + 3 \)[\( \because d = 16 - a \)]
\( \Rightarrow 3a = 32 - 2a + 3 \)
\( \Rightarrow 5a = 35 \)
\( \Rightarrow a = 7 \)
Hence, \( d = 16 - 7 = 9 \)
Hence, the required A.P. is 7, 16, 25,…..

Question. Which term of the A.P. 3, 15, 27, 39, ..... will be 120 more than its \( 21^{st} \) term ?*
Answer: The given A.P. is 3, 15, 27, 39, .....
Here \( a = 3, d = 12 \)
\( \therefore a_{21} = a + 20d \)
\( = 3 + 20 \times 12 \)
\( = 3 + 240 = 243 \)
Now, \( a_n = a_{21} + 120 \)
\( = 243 + 120 = 363 \)
\( a_n = a + (n - 1) \times 12 \)
\( \Rightarrow 363 = 3 + (n - 1) 12 \)
or \( 360 = (n - 1) \times 12 \)
or \( n - 1 = 30 \)
\( n = 31 \)
Hence, the term which is 120 more than its \( 21^{st} \) term will be its \( 31^{st} \) term.

Question. Find the sum of all multiples of 7 lying between 500 and 900.*
Answer: Multiples of 7 lying between 500 and 900 are
504, 511, 518, 525, 532, …, 896
We observe that the above series is an A.P. with
\( a = 504, d = 7 \) and \( T_n = 896 \)
Thus, \( T_n = 504 + (n - 1)7 = 896 \)
\( \Rightarrow 7(n - 1) = 392 \)
\( \Rightarrow n - 1 = 56 \)
\( \Rightarrow n = 57 \)
Thus, \( S_{57} = \frac{57}{2} \{2(504) + (57 - 1)7\} \)
\( = \frac{57}{2} \{1008 + 392\} \)
\( = \frac{57}{2} (1400) \)
\( = 39900 \).

Question. Find the sum of the first \( n \) terms of an A.P. whose \( n^{th} \) term is \( 5n - 1 \). Find the sum of the first 20 terms.*
Answer: Given, \( T_n = 5n - 1 \)
We know that, \( T_n = a + (n - 1)d \)
Thus, \( a + (n - 1)d = 5n - 1 \)
\( \Rightarrow a + nd - d = 5n - 1 \)
\( \Rightarrow nd + (a - d) = 5n - 1 \)
On comparing both sides, we get
\( \Rightarrow d = 5 \) and \( a - d = - 1 \)
\( \Rightarrow a = 5 - 1 = 4 \)
Thus, \( S_{20} = \frac{20}{2} \{2(4) + (20 - 1)5\} \)
\( = 10(8 + 95) \)
\( = 1030 \).

Question. In a given A.P., if the \( p^{th} \) term is \( q \) and the \( q^{th} \) term is \( p \), then show that the \( n^{th} \) term is \( ( p + q - n) \).**
Answer: Let the first term be \( a \) and the common difference be \( d \).
Thus, \( T_p = a + ( p - 1)d = q \)
\( \Rightarrow a - d = q - pd \)…(i)
and \( T_q = a + (q - 1)d = p \)
\( \Rightarrow a - d = p - qd \)…(ii)
From equations (i) and (ii),
\( q - pd = p - qd \)
\( \Rightarrow d(q - p) = p - q \)
\( \Rightarrow d = - 1 \)
and \( a = p + q - 1 = q + p - 1 \)
So, \( T_n = \{a + (n - 1)d\} \)
\( = p + q - 1 + (n - 1) (- 1) \)
\( = p + q - 1 - n + 1 \)
\( = p + q - n \). Hence Proved.

Question. If the \( 10^{th} \) term of an A.P. is 52 and the \( 17^{th} \) term is 20 more than the \( 13^{th} \) term, find the A.P.
Answer: Let \( a \) be the first term and \( d \) be the common difference.
So, \( T_{10} = a + (10 - 1)d = 52 \)[Given]
\( \Rightarrow a + 9d = 52 \)…(i)
Now, \( T_{17} = T_{13} + 20 \)[Given]
\( \Rightarrow a + (17 - 1)d = a + (13 - 1)d + 20 \)
\( a + 16d = a + 12d + 20 \)
\( \Rightarrow 16d = 12d + 20 \)
\( \Rightarrow 4d = 20 \)
\( \Rightarrow d = 5 \)
From equation (i),
\( \Rightarrow a = 52 - 9(5) \)
\( = 52 - 45 = 7 \)
Hence, the required A.P. is 7, 12, 17, 22, …

Question. The sum of the three numbers in A.P. is 21 and their product is 231. Find the numbers.
Answer: Let the numbers be \( (a - d ), a, (a + d ) \).
So, the first term = \( a - d \)
and the common difference = \( d \)
We know, \( S_n = \frac{n}{2} \{2a + (n - 1) d\} \)
\( \Rightarrow S_3 = \frac{3}{2} \{2(a - d) + (3 - 1)d\} \)
\( \Rightarrow 21 = \frac{3}{2} \{2a - 2d + 2d\} \)[Given]
\( \Rightarrow 3a = 21 \)
\( \Rightarrow a = 7 \)
Now, \( (a - d ) a(a + d ) = 231 \)[Given]
\( \Rightarrow (7 - d ) 7(7 + d ) = 231 \)
\( \Rightarrow 49 - d^2 = 33 \)
\( \Rightarrow d^2 = 16 \)
\( \Rightarrow d = \pm 4 \)
Thus, the numbers are either 3, 7, 11 or 11, 7, 3.

Question. Find the sum of all two-digit odd positive numbers.
Answer: Two-digit odd positive integers are, 11, 13, 15, ............ 99.
The above series is an A.P. with
\( a = 11, d = 2 \) and \( T_n = 99 \)
Now, \( T_n = 11 + (n - 1)2 = 99 \)
\( \Rightarrow (n - 1)2 = 88 \)
\( \Rightarrow n - 1 = 44 \)
\( \Rightarrow n = 45 \)
\( \therefore S_{45} = \frac{45}{2} \{2(11) + (45 - 1)(2)\} \)
\( = \frac{45}{2} \{2(11) + (44)(2)\} \)
\( = 45\{11 + 44\} \)
\( = 45 \times 55 \)
\( = 2475 \).

Question. The \( 16^{th} \) term of an A.P. is five times its third term. If its \( 10^{th} \) term is 41, then find the sum of its first fifteen terms.*
Answer: Given that \( 16^{th} \) term of an A.P. is five time its third term.
\( \therefore a + (16 - 1)d = 5[a + (3 - 1)d] \)
\( \Rightarrow a + 15d = 5[a + 2d] \)
\( \Rightarrow a + 15d = 5a + 10d \)
\( \Rightarrow 4a - 5d = 0 \)…(i)
Also given that,
\( T_{10} = 41 \)
\( \Rightarrow a + (10 - 1)d = 41 \)
\( \Rightarrow a + 9d = 41 \)…(ii)
On multiplying equation (ii) by 4, we get
\( 4a + 36d = 164 \)…(iii)
Subtracting equation (iii) from (i), we get
\( - 41d = - 164 \)
\( \Rightarrow d = 4 \)
On substituting the value of \( d \) in (i), we get
\( 4a - 5 \times 4 = 0 \)
\( \Rightarrow 4a = 20 \)
\( \Rightarrow a = 5 \)
Now, \( S_{15} = \frac{15}{2} [2a + (15 - 1)d ] \)
\( S_{15} = \frac{15}{2} (2 \times 5 + 14 \times 4) \)
\( = \frac{15}{2} \times 2(5 + 14 \times 2) \)
\( = 15(5 + 28) = 15 \times 33 \)
\( S_{15} = 495 \)
Hence, sum of first fifteen terms is 495.

Question. How many terms of the A.P. 3, 5, 7, 9, … must be added to get the sum of 120 ?
Answer: Given A.P. is, 3, 5, 7, 9, .....
Thus, \( a = 3 \) and \( d = 2 \)
Now, \( S_n = \frac{n}{2} \{2(3) + (n - 1) (2)\} \)
\( \Rightarrow n\{3 + (n - 1)\} = 120 \)
\( \Rightarrow n(n + 2) = 120 \)
\( \Rightarrow n^2 + 2n - 120 = 0 \)
\( \Rightarrow n^2 + 12n - 10n - 120 = 0 \)
\( \Rightarrow n(n + 12) - 10(n + 12) = 0 \)
\( \Rightarrow (n + 12) (n - 10) = 0 \)
\( \Rightarrow n = 10 \) or \( - 12 \)
Here, the given A.P. series is increasing, so \( n \) cannot be negative.
Hence, so the required answer is 10.

Question. If the sum of the first \( n \) terms of an A.P. is given by \( S_n = \frac{n}{2} (3n + 5) \), then find its \( 25^{th} \) term.
Answer: Let the first term be \( a \) and the common difference be \( d \).
It is given that,
\( S_n = \frac{n}{2} \{2a + (n - 1)d\} = \frac{n}{2} (3n + 5) \)
\( \Rightarrow 2a + (n - 1)d = 3n + 5 \)
\( \Rightarrow (2a - d ) + dn = 5 + 3n \)
On comparing, we get
\( d = 3 \) and \( 2a - d = 5 \)
\( \Rightarrow 2a = 8 \)
\( \Rightarrow a = 4 \)
\( \therefore T_{25} = 4 + (25 - 1)3 \)
\( = 4 + 24 \times 3 \)
\( = 76 \).

Question. The \( 14^{th} \) term of an A.P. is twice its \( 8^{th} \) term. If its 6th term is \( -8 \), then find the sum of its first 20 terms.*
Answer: In the given AP, let first term = \( a \) and common difference = \( d \)
Then, \( T_n = a + (n - 1)d \)
\( \Rightarrow T_{14} = a + (14 - 1)d = a + 13d \)
and \( T_8 = a + (8 - 1)d = a + 7d \)
Now, it is given that,
\( T_{14} = 2T_8 \)
\( \Rightarrow a + 13d = 2(a + 7d ) \)
\( \Rightarrow a + 13d = 2a + 14d \)
\( a = - d \)...(i)
Also, \( T_6 = a + (6 - 1)d \)
\( \Rightarrow a + 5d = - 8 \)...(ii)
Substituting the value of \( a \) from equation (i) and equation (ii), we get
\( - d + 5d = - 8 \)
\( 4d = - 8 \)
\( d = - 2 \)
From equation (i),
\( a = - d = - (- 2) = 2 \)
\( \therefore \) Sum of first 20 terms is
\( S_{20} = \frac{20}{2} \{2a + (n - 1)d \} \)
\( = 10 \{2 \times 2 + (20 - 1)(-2)\} \)
\( = 10\{4 - 38\} \)
\( = -340 \).

Question. The sum of the first 7 terms of an A.P. is 49 and the sum of the first 17 terms is 289. Find the sum of first \( n \) terms.** [NCERT]
Answer: Let the first terms be \( a \) and the common difference be \( d \).
Now, it is given that,
\( S_7 = \frac{7}{2} \{2a + (7 - 1)d\} = 49 \)
\( \Rightarrow 49 = \frac{7}{2} \{2a + 6d\} \)
\( \Rightarrow a + 3d = 7 \)…(i)
and \( S_{17} = \frac{17}{2} \{2a + (17 - 1)d\} = 289 \)
\( \Rightarrow 289 = \frac{17}{2} \{2a + (16)d\} \)
\( \Rightarrow a + 8d = 17 \)…(ii)
Subtracting equation (i) from equation (ii), we have
\( 5d = 10 \)
\( \Rightarrow d = 2 \)
From equation (i),
\( a = 7 - 3d \)
\( = 7 - 3(2) \)
\( = 7 - 6 = 1 \)
\( \therefore S_n = \frac{n}{2} \{2(1) + (n - 1)2\} \)
\( \Rightarrow S_n = \frac{n}{2} \{2 + 2n - 2\} = n^2 \)
Hence, the sum of \( n \) terms will be \( n^2 \).

Question. If the sum of first \( n \) terms of an A.P. is \( n^2 \), then find its \( 10^{th} \) term.*
Answer: Given, \( S_n = n^2 \), \( S_{n - 1} = (n - 1)^2 \)
\( a_n = S_n - S_{n - 1} = n^2 - (n - 1)^2 \)
\( = n^2 - [n^2 - 2n + 1] \)
\( = n^2 - n^2 + 2n - 1 \)
\( a_n = 2n - 1 \)
Put \( n = 10 \),
\( \therefore a_{10} = 2 \times 10 - 1 = 19 \)
Hence \( 10^{th} \) term = 19

Question. The \( 9^{th} \) term of an A.P. is equal to six times its \( 2^{nd} \) term. If its \( 5^{th} \) term is 22, find the A.P.
Answer: Let the first term be \( a \) and the common difference be \( d \).
Given, \( T_9 = 6T_2 \)
\( \Rightarrow \{a + (9 - 1)d\} = 6\{a + (2 - 1)d\} \)
\( \Rightarrow a + 8d = 6a + 6d \)
\( \Rightarrow 5a = 2d \)…(i)
and \( T_5 = \{a + (5 - 1)d\} = 22 \)
\( \Rightarrow a + 4d = 22 \)
\( \Rightarrow a + 2 (2d) = 22 \)
\( \Rightarrow a + 10a = 22 \) [\( \because 5a = 2d \)]
\( \Rightarrow 11a = 22 \)
\( \Rightarrow a = 2 \)
Substituting \( a = 2 \) in equation (i), we get
\( 10 = 2d \)
\( \Rightarrow d = 5 \)
Hence, the required A.P. is 2, 7, 12, 17, …

Question. The \( 24^{th} \) term of an A.P. is twice its \( 10^{th} \) term. Show that its \( 72^{nd} \) term is 4 times its \( 15^{th} \) term.
Answer: Let the first term be \( a \) and the common difference be \( d \).
Given, \( T_{24} = 2T_{10} \)
\( \Rightarrow \{a + (24 - 1)d\} = 2\{a + (10 - 1)d\} \)
\( \Rightarrow a + 23d = 2a + 18d \)
\( \Rightarrow a = 5d \)…(i)
Now, \( T_{15} = \{a + (15 - 1)d\} \)
\( \Rightarrow T_{15} = a + 14d \)
\( \Rightarrow T_{15} = 5d + 14d \) [\( \because a = 5d \)]
\( \Rightarrow T_{15} = 19d \)…(ii)
and \( T_{72} = a + (72 - 1)d \)
\( \Rightarrow T_{72} = a + 71d \)
\( \Rightarrow T_{72} = 5d + 71d \)
\( \Rightarrow T_{72} = 76d = 4(19d) \)
\( \Rightarrow T_{72} = 4 T_{15} \) [from (ii)]
Hence Proved.

Question. The digits of a positive number of three digits are in A.P. and their sum is 15. The number obtained by reversing the digits is 594 less than the original number. Find the number.*
Answer: Let the three digits of a positive number be \( a - d, a, a + d \)
\( \therefore a - d + a + a + d = 15 \)[Given]
\( \Rightarrow 3a = 15 \)
\( \Rightarrow a = 5 \)
Original number = \( 100 (a - d ) + 10 (a) + a + d \)
\( = 100a - 100d + 10a + a + d \)
\( = 111a - 99d \)
and number obtained by reversing the digits
\( = 100 (a + d ) + 10 (a) + a - d \)
\( = 100a + 100d + 10a + a - d \)
\( = 111a + 99d \)
Now, original no. - Reversed no. = 594
\( 111a - 99d - (111a + 99d) = 594 \)
\( \Rightarrow -198d = 594 \)
\( \Rightarrow d = -3 \)
\( \therefore \) Original number = \( 111(5) - 99(-3) = 555 + 297 = 852 \).

Question. If the sum of the first \( n \) terms of an A.P. is given by \( S_n = \frac{n}{2} (3n + 13) \), then find its \( 25^{th} \) term.
Answer: Let the first term be \( a \) and the common difference be \( d \).
Given, \( S_n = \frac{n}{2} \{2a + (n - 1)d\} = \frac{n}{2} (3n + 13) \)
\( \Rightarrow 2a + (n - 1)d = 3n + 13 \)
\( \Rightarrow (2a - d ) + dn = 13 + 3n \)
On comparing, we get
\( d = 3 \) and \( 2a - d = 13 \)
\( \therefore 2a = 16 \)
\( \Rightarrow a = 8 \)
\( \therefore T_{25} = 8 + (25 - 1)3 \)
\( T_{25} = 8 + 24 \times 3 = 80 \).

Question. If \( S_n \), denotes the sum of first \( n \) terms of an A.P., prove that \( S_{12} = 3(S_8 - S_4) \).
Answer: Let be the first term be \( a \) and the common difference be \( d \),
We know, \( S_n = \frac{n}{2} [2a + (n - 1)d] \)
\( \therefore S_{12} = \frac{12}{2} [2a + (12 - 1)d ] \)
\( = 6 (2a + 11d ) \)
\( = 12a + 66d \)...(i)
\( S_8 = \frac{8}{2} [2a + (8 - 1)d ] \)
\( = 4(2a + 7d) \)
\( = 8a + 28d \)...(ii)
and \( S_4 = \frac{4}{2} [2a + (4 - 1)d] \)
\( = 2(2a + 3d) = 4a + 6d \)...(iii)
Given, \( 3(S_8 - S_4) = 3(8a + 28d - 4a - 6d) \) [from (ii) and (iii)]
\( = 3(4a + 22d) \)
\( = 12a + 66d \)
\( = S_{12} \)[from (i)]
Hence Proved.

Question. Pradeep repays the total loan of ₹ 1, 18000 by paying every month starting with the first instalment of ₹ 1000. He increases the instalment by ₹ 100 every month. What amount will he pay as the last instalment of loan ?
Answer: Ist instalment = ₹ 1000
IInd instalment = ₹ (1000 + 100) = ₹ 1100
IIIrd instalment = ₹ (1100 + 100) = ₹ 1200
Let the number of instalments be \( n \).
\( \therefore \) Sum \( = 1000 + 1100 + 1200 + \dots + n^{th} \) term
These terms are in AP with \( a = 1000, d = 100 \).
\( \therefore \) Sum \( = \frac{n}{2} [ 2 \times 1000 + (n - 1)100] = 118000 \)
\( \Rightarrow \frac{n}{2} (2000 + 100n - 100) = 118000 \)
\( \Rightarrow n (1900 + 100n) = 236000 \)
\( \Rightarrow 100n^2 + 1900n - 236000 = 0 \)
\( \Rightarrow n^2 + 19n - 2360 = 0 \)
\( \Rightarrow n^2 + 59n - 40n - 2360 = 0 \)
\( \Rightarrow (n + 59) (n - 40) = 0 \)
\( \Rightarrow n = 40 \) or \( n = - 59 \) (Rejected)
\( \therefore \) Last instalment, \( a_{40} = a + 39d \)
\( = 1000 + 39 \times 100 \)
\( = Rs 4900 \)

Question. Saurav gets pocket money from his father every day. Out of the pocket money, he saves ₹ 2.75 on first day and on each succeeding day he increases his saving by 25 paise. Find
(i) the amount saved by Saurav on 14th day,
(ii) the amount saved by Saurav on 25th day,
(iii) the total amount saved by Saurav in 30 days.
Answer: Money saved on Ist day = Rs 2.75
Money saved on IInd day = Rs (2.75 + 0.25) = Rs 3.00
Money saved on IIIrd day = Rs (3.00 + 0.25) = Rs 3.25.
Money saved by Saurav forms an AP with \( a = 2.75 \) and \( d = 0.25 \)
(i) Money saved on 14th day \( = a + 13d \)
\( = 2.75 + 13 \times 0.25 = 2.75 + 3.25 = Rs 6.00 \)
(ii) Money saved on 25th day \( = a + 24d \)
\( = 2.75 + 24 \times 0.25 = Rs 8.75 \)
(iii) Total amount of money saved in 30 days
\( = \frac{30}{2} (2 \times 2.75 + 29 \times 0.25) = Rs 191.25 \)

Question. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studing, e.g., a section of class I will plant 1 tree, a section of class II will plant 2 trees and so on till class XII. There are three sections of each class. How many trees will be planted by the students ?
Answer: Number of trees planted by students of class I \( = 3 \times 1 = 3 \)
Number of trees planted by students of class II \( = 3 \times 2 = 6 \)
Number of trees planted by students of class III \( = 3 \times 3 = 9 \)
Number of trees planted by students of class XII \( = 3 \times 12 = 36 \)
Total number of trees planted \( = 3 + 6 + 9 + \dots + 36 \)
These are 12 terms of an A.P. with \( a = 3 \) and last term \( = 36 \)
\( \Rightarrow l = 36 \)
\( \therefore \) Sum \( = \frac{n}{2} (a + l ) = \frac{12}{2} (3 + 36 ) = 234 \).

Question. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line. Each competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run ?
Answer: We have,
\( d_1 = \) Distance run by the competitor to pick up first potato \( = 2 \times 5 \) m
\( d_2 = \) Distance run by the competitor to pick up second potato \( = 2(5 + 3) \) m
\( d_3 = \) Distance run by the competitor to pick up third potato \( = 2(5 + 2 \times 3) \) m
\( d_4 = \) Distance run by the competitor to pick up fourth potato \( = 2(5 + 3 \times 3) \) m
\( d_n = \) Distance run by the competitor to pick up \( n^{th} \) potato \( = 2\{5 + (n - 1) \times 3\} \) m
Total distance run by the competitor to pick up potatoes
\( = d_1 + d_2 + d_3 + \dots + d_n \)
\( = 2 \times 5 + 2(5 + 3) + 2(5 + 2 \times 3) + 2 (5 + 3 \times 3) + \dots + 2 \{5 + (n - 1) \times 3\} \) metres
\( = 2[5 + \{5 + 3\} + \{5 + (2 \times 3)\} + \{5 + (3 \times 3)\} + \dots + \{ 5 + (n - 1) \times 3\}] \)
\( = 2[(5 + 5 + \dots + 5) + \{3 + (2 \times 3) + (3 \times 3) + \dots + (n - 1) \times 3\}] \)
\( = 2 [(5 \times n) + 3 \{1 + 2 + 3 + \dots + (n - 1)\}] \)
\( = 2 \left[ 5n + 3 \left( \frac{n-1}{2} \right) \{1 + (n - 1)\} \right] \)
\( = 2 \left[ 5n + \frac{3n(n-1)}{2} \right] \)
\( = [10n + 3n (n - 1)] \)
\( = 3n^2 + 7n \)
\( = n (3n + 7) \) metres
Since, \( n = 10 \)
\( \therefore \) Total distance run by the competitor \( = 10(3(10) + 7) = 370 \) m.

Question. Shyam was late by 5 minutes in joining his duty on the first working day. On the second day, he was late by 10 minutes, on third day by 15 minutes and so on. After 25 working days he was shunted out of the job. Find the total working time avoided by Shyam.
Answer: On first day, Shyam was late by 5 minutes
On second day, he was late by 10 minutes
On third day, he was late by 15 minutes and so on
The time by which Shyam was late for his office forms an A.P.
\( \therefore a = 5, d = 10 - 5 = 5, n = 25 \)
So, total working time avoided by Shyam,
\( S_{25} = \frac{25}{2} \{2a + (25 - 1) d\} \)
\( = \frac{25}{2} \{2 \times 5 + (25 - 1) 5\} \)
\( = \frac{25}{2} \{2 \times 5 + 24 \times 5\} \)
\( = 25 \{5 + 12 \times 5\} \)
\( = 25 \{5 + 60\} \)
\( = 25 \times 65 = 1625 \) minutes

Long Answer Type Questions

Question. In an A.P. of 50 terms, the sum of the first 10 terms is 210 and the sum of the last 15 terms is 2565. Find the A.P.
Answer: Let \( a \) be the first term and \( d \) be the common difference.
Now, \( S_{10} = \frac{10}{2} [2a + (10 - 1)d ] \)
\( \Rightarrow 210 = 5(2a + 9d) \)
\( \Rightarrow 2a + 9d = 42 \)…(i)
and \( 15^{th} \) term from the end is \( (50 - 15 + 1) = 36^{th} \) term from the beginning.
So, \( t_{36} = a + (36 - 1)d = a + 35d \)
Hence, the sum of the last 15 terms
\( = \frac{15}{2} \{2t_{36} + (15 - 1)d \} \)
\( \Rightarrow 2565 = \frac{15}{2} \{2(a + 35d ) + 14d\} \)
\( \Rightarrow 2565 = 15\{a + 35d + 7d\} \)
\( \Rightarrow a + 42d = 171 \)
\( \Rightarrow 2a + 84d = 342 \)…(ii)
Subtracting (i) from (ii), we have
\( 75d = 300 \Rightarrow d = 4 \)
From (i), \( 2a + 9(4) = 42 \)
\( \Rightarrow 2a = 42 - 36 = 6 \)
\( \Rightarrow a = 3 \)
Thus, the A.P. is 3, 7, 11, … 199.

Question. If \( S_n \) denotes the sum of the first \( n \) terms of an A.P., prove that \( S_{30} = 3(S_{20} - S_{10}) \).
Answer: Let \( a \) be the first term of the series and \( d \) be the common difference.
\( \therefore S_n = \frac{n}{2} \{2a + (n - 1)d\} \)
So, \( S_{30} = \frac{30}{2} \{2a + (30 - 1)d\} = 15\{2a + 29d] \)…(i)
\( S_{20} = \frac{20}{2} \{2a + (20 - 1)d\} = 10\{2a + 19d\} \)…(ii)
and \( S_{10} = \frac{10}{2} \{2a + (10 - 1)d\} = 5\{2a + 9d\} \)…(iii)
Now, \( S_{20} - S_{10} = 10\{2a + 19d\} - 5\{2a + 9d\} \)
\( = 20a + 190d - 10a - 45d \)
\( = 10a + 145d \)
\( = 5(2a + 29d ) \) [From (iii)]
\( 3(S_{20} - S_{10}) = 3[5 (2a + 29d] \)
\( S_{30} = 15 (2a + 29d) \) [From (i)]
Thus, \( 3(S_{20} - S_{10}) = S_{30} \). Hence Proved.

Question. Find the sum of the middlemost terms of the A.P. \( -\frac{4}{3}, -1, -\frac{2}{3}, \dots, 4\frac{1}{3} \).
Answer: The given A.P. is \( -\frac{4}{3}, -1, -\frac{2}{3}, \dots, \frac{13}{3} \)
Thus, \( a = -\frac{4}{3} \), \( d = -1 - (-\frac{4}{3}) = -1 + \frac{4}{3} = \frac{1}{3} \)
and \( t_n = 4\frac{1}{3} = \frac{13}{3} \)
\( t_n = a + (n - 1)d \)
\( \Rightarrow \frac{13}{3} = -\frac{4}{3} + (n - 1)\frac{1}{3} \)
\( \Rightarrow \frac{13}{3} = \frac{-4 + n - 1}{3} \)
\( \Rightarrow \frac{13}{3} = \frac{-5 + n}{3} \)
\( \Rightarrow n = 18 \)
Thus, the middlemost terms of the A.P. are the \( 9^{th} \) and \( 10^{th} \) terms.
So \( t_9 = a + (9 - 1)d = a + 8d = -\frac{4}{3} + \frac{8}{3} = \frac{4}{3} \)
and \( t_{10} = a + (10 - 1)d = a + 9d = -\frac{4}{3} + \frac{9}{3} = \frac{5}{3} \)
Thus, \( t_9 + t_{10} = \frac{4}{3} + \frac{5}{3} = \frac{9}{3} = 3 \).

Question. Find the common difference of an A.P. whose first term is 5 and the sum of the first 4 terms is half of the sum of the next four terms.
Answer: Given, \( a = 5 \)
Thus, \( S_4 = \frac{4}{2} \{2a + (4 - 1)d\} = 2( 2a + 3d ) \)
\( t_5 = \{a + (5 - 1)d\} = (a + 4d ) \)
For the next set of 4 numbers, \( a = t_5 \) and \( d \) remains the same
Thus, \( S'_4 = \frac{4}{2} \{2(a + 4d ) + (4 - 1)d\} \)
\( = 2(2a + 8d + 3d) = 2(2a + 11d ) \)
According to the question,
\( S_4 = \frac{1}{2} S'_4 \)
\( \Rightarrow 2(2a + 3d ) = \frac{1}{2} \times 2(2a + 11d) \)
\( \Rightarrow 2(2a + 3d ) = 2a + 11d \)
\( \Rightarrow 4a + 6d = 2a + 11d \)
\( \Rightarrow 2a = 5d \)
\( \Rightarrow 2(5) = 5d \Rightarrow 10 = 5d \Rightarrow d = 2 \)
Thus, the common difference is 2.

Question. The sum of the \( 4^{th} \) and \( 8^{th} \) terms of an A.P. is 24 and the sum of the \( 6^{th} \) and \( 10^{th} \) terms is 44. Find the first three terms of the A.P.
Answer: Let the first term be \( a \), the common difference be \( d \) and the number of terms be \( n \).
Thus, \( t_4 = a + (4 - 1)d = a + 3d \)
\( t_8 = a + (8 - 1)d = a + 7d \)
\( t_6 = a + (6 - 1)d = a + 5d \)
\( t_{10} = a + (10 - 1)d = a + 9d \)
Now, \( t_4 + t_8 = 24 \)
\( \Rightarrow a + 3d + a + 7d = 24 \Rightarrow 2a + 10d = 24 \Rightarrow a + 5d = 12 \)…(i)
and \( t_6 + t_{10} = 44 \)
\( \Rightarrow a + 5d + a + 9d = 44 \Rightarrow 2a + 14d = 44 \Rightarrow a + 7d = 22 \)…(ii)
Subtracting (i) from (ii), we have
\( 2d = 10 \Rightarrow d = 5 \)
From (i) \( a = 12 - 5(5) = 12 - 25 = - 13 \)
Hence, first three terms of the A.P. are – 13, – 8, – 3.

Question. The sum of the first 6 terms of an A.P. is 42. If its \( 10^{th} \) and \( 30^{th} \) terms are in the ratio 1 : 3, find the \( 1^{st} \) and \( 13^{th} \) terms.
Answer: Let \( a \) be the first term and \( d \) be the common difference of the A.P.
So, \( S_6 = \frac{6}{2} \{2a + (6 - 1)d\} \)
\( \Rightarrow 42 = 3(2a + 5d ) \Rightarrow 2a + 5d = 14 \)…(i)
\( t_{10} = a + (10 - 1)d = a + 9d \)
\( t_{30} = a + (30 - 1)d = a + 29d \)
Now \( t_{10} : t_{30} = 1 : 3 \)
\( \Rightarrow \frac{a + 9d}{a + 29d} = \frac{1}{3} \)
\( \Rightarrow 3a + 27d = a + 29d \)
\( \Rightarrow 2a = 2d \Rightarrow a = d \)…(ii)
Substituting (ii) in (i),
\( 2a + 5a = 14 \Rightarrow 7a = 14 \Rightarrow a = 2 \)
Thus, \( d = 2 \)
Now \( t_{13} = 2 + (13 - 1)2 = 2 + 24 = 26 \)
So the first term is 2 and the \( 13^{th} \) term is 26.

Question. If the ratio of the sum of the first \( n \) terms of two A.Ps. is \( (7n + 1) : (4n + 27) \), then find the ratio of their 9th terms.
Answer: Let the first terms be \( a \) and \( a' \) and \( d \) and \( d' \) be their respective common differences.
Then, \( \frac{S_n}{S'_n} = \frac{\frac{n}{2}[2a+(n-1)d]}{\frac{n}{2}[2a'+(n-1)d']} = \frac{7n+1}{4n+27} \)
\( \Rightarrow \frac{a + (\frac{n-1}{2})d}{a' + (\frac{n-1}{2})d'} = \frac{7n+1}{4n+27} \)
To get ratio of \( 9^{th} \) terms, replacing \( \frac{n-1}{2} = 8 \)
\( \Rightarrow n = 17 \)
Hence \( \frac{a + 8d}{a' + 8d'} = \frac{7 \times 17 + 1}{4 \times 17 + 27} = \frac{120}{95} = \frac{24}{19} \).

Question. Find the number of terms of the A.P. 64, 60, 56, … so that sum is 544. Explain the double answer.
Answer: Given A.P. is 64, 60, 56, .....
So, \( a = 64, d = - 4 \) and \( S_n = 544 \)
Let the number of terms be \( n \).
Thus, \( S_n = \frac{n}{2} \{2(64) + (n - 1)(- 4)\} = 544 \)
\( \Rightarrow n \{64 + (n - 1)(- 2)\} = 544 \)
\( \Rightarrow n \{66 - 2n\} = 544 \)
\( \Rightarrow n \{33 - n\} = 272 \)
\( \Rightarrow 33n - n^2 = 272 \)
\( \Rightarrow n^2 - 33n + 272 = 0 \)
\( \Rightarrow n^2 - 17n - 16n + 272 = 0 \)
\( \Rightarrow n(n - 17) - 16(n - 17) = 0 \)
\( \Rightarrow (n - 17) (n - 16) = 0 \)
\( \Rightarrow n = 16 \) or \( 17 \)
The double answer is because of the zero. It does not add to the total sum but is definitely a number in the series (\( 17^{th} \) term).
\( t_{17} = 64 + (17 - 1)(- 4) = 0 \)
\( i.e. \), \( 17^{th} \) term is zero. So sum of all 16 and 17 terms is zero as 0 will not contribute to the sum.

Question. Find the sum of the following : \( (1 - \frac{1}{n}) + (1 - \frac{2}{n}) + (1 - \frac{3}{n}) + \dots \) upto \( n \) terms.
Answer: Given,
\( (1 - \frac{1}{n}) + (1 - \frac{2}{n}) + (1 - \frac{3}{n}) + \dots + (1 - \frac{n}{n}) \)
\( = (1 + 1 + 1 + \dots + 1) - \left( \frac{1}{n} + \frac{2}{n} + \frac{3}{n} + \dots + \frac{n}{n} \right) \)
\( = n - \left[ \frac{n}{2} \{ 2(\frac{1}{n}) + (n - 1)\frac{1}{n} \} \right] \)
\( = n - \left[ \frac{n}{2} \{ \frac{2}{n} - \frac{1}{n} + 1 \} \right] \)
\( = n - \left[ \frac{n}{2} \{ \frac{1}{n} + 1 \} \right] \)
\( = n - \frac{1+n}{2} = \frac{2n - n - 1}{2} = \frac{n-1}{2} \).
Thus, the sum is \( \frac{n - 1}{2} \).

Question. If \( m^{th} \) term of an A.P. is \( \frac{1}{n} \) and \( n^{th} \) term is \( \frac{1}{m} \), then find the sum of its first \( mn \) terms.
Answer: Let \( a \) and \( d \) be the first term and common difference respectively of the given A.P.
Then, \( \frac{1}{n} = a + (m - 1) d \)…(i)
\( \frac{1}{m} = a + (n - 1) d \)…(ii)
Subtracting equation (ii) from equation (i),
\( \frac{1}{n} - \frac{1}{m} = (m - n) d \)
\( \Rightarrow \frac{m - n}{nm} = (m - n) d \)
\( \Rightarrow d = \frac{1}{mn} \)
Putting \( d = \frac{1}{mn} \) in equation (i), we get \( a = \frac{1}{mn} \).
\( S_{mn} = \frac{mn}{2} [ 2(\frac{1}{mn}) + (mn - 1)(\frac{1}{mn}) ] \)
\( = \frac{mn}{2} \left[ \frac{2 + mn - 1}{mn} \right] = \frac{mn + 1}{2} \).

Question. The houses in a row are numbered consecutively from 1 to 49. Show that there exists a value of \( X \) such that sum of numbers of houses proceeding the house numbered \( X \) is equal to sum of the numbers of houses following \( X \).*
Answer: Given, the houses in a row numbered consecutively from 1 to 49.
Now, sum of numbers preceding the number \( X \)
\( = \frac{X(X - 1)}{2} \)
and sum of numbers following the number \( X \)
\( = \frac{49(50)}{2} - \frac{X(X - 1)}{2} - X \)
\( = \frac{2450 - X^2 + X - 2X}{2} \)
\( = \frac{2450 - X^2 - X}{2} \)
According to the question,
Sum of no’s preceding \( X \) = Sum of no’s following \( X \)
\( \frac{X(X - 1)}{2} = \frac{2450 - X^2 - X}{2} \)
\( \Rightarrow X^2 - X = 2450 - X^2 - X \)
\( \Rightarrow 2X^2 = 2450 \)
\( \Rightarrow X^2 = 1225 \)
\( \Rightarrow X = 35 \)
Hence, at \( X = 35 \), sum of number of houses preceeding the house no. \( X \) is equal to sum of the number of houses following \( X \).

Question. A thief runs with a uniform speed of 100 m/minute. After one minute, a policeman runs after the thief to catch him. He goes with a speed of 100 m/minute in the first minute and increases his speed by 10 m/minute every succeeding minute. After how many minutes the policeman will catch the thief ?*
Answer: Let total time be \( n \) minutes.
Since, policeman runs after 1 minute so he will catch the thief in \( (n – 1) \) minutes.
Total distance covered by thief
= 100 m/minute × \( n \) minute
= \( (100 n) \) m
Now, total distance covered by the policeman
= \( (100) \)m + \( (100 + 10) \)m + \( (100 + 10 + 10) \)m + ......+ \( (n – 1) \) terms
\( i.e. \), 100 + 110 + 120 + ..... + \( (n – 1) \) terms
\( \therefore S_{n - 1} = \frac{n-1}{2} [2 \times 100 + (n - 2) 10] \)
Now, \( \frac{n - 1}{2} [200 + (n – 2) 10] = 100 n \).
\( \Rightarrow (n – 1) (200 + 10n – 20) = 200 n \).
\( \Rightarrow 200 n – 200 + 10n^2 – 10n + 20 – 20n = 200n \)
\( \Rightarrow 10n^2 – 30n – 180 = 0 \)
\( \Rightarrow n^2 – 3n – 18 = 0 \)
\( \Rightarrow n^2 – (6 – 3) n – 18 = 0 \)
\( \Rightarrow n^2 – 6n + 3n – 18 = 0 \)
\( \Rightarrow n(n – 6) + 3(n – 6) = 0 \)
\( \Rightarrow (n + 3) (n – 6) = 0 \)
\( \therefore n = 6 \) or \( n = – 3 \) (neglect)
Hence, policeman will catch the thief in \( (6 – 1) \) \( i.e. \), 5 minutes.