CBSE Class 10 Mathematics Arithmetic Progressions Worksheet Set 14

Question. The sum of the first 15 multiples of 8 is 
(a) 900
(b) 960
(c) 1000
(d) 870
Answer: (b) 960
Explanation: Multiples of 8 are 8, 16, 24, .........
Here \( a = 8, d = 16 - 8 = 8 \) and \( n = 15 \)
Now, \( S_n = \frac{n}{2} [2a + (n - 1)d] \)

\( \implies S_{15} = \frac{15}{2} [2 \times 8 + (15 - 1)8] \)

\( \implies S_{15} = \frac{15}{2} [16 + 14 \times 8] \)

\( \implies S_{15} = \frac{15}{2} \times 128 \)

\( \implies S_{15} = 15 \times 64 \)

\( \implies S_{15} = 960 \)

Question. If the angles of a right angled triangle are in A.P. then the angles of that triangle will be
(a) \( 45^\circ, 45^\circ, 90^\circ \)
(b) \( 30^\circ, 60^\circ, 90^\circ \)
(c) \( 40^\circ, 50^\circ, 90^\circ \)
(d) \( 20^\circ, 70^\circ, 90^\circ \)
Answer: (b) \( 30^\circ, 60^\circ, 90^\circ \)
Explanation: Let the three angles of a triangle be \( a - d, a \) and \( a + d \).
\( \therefore a - d + a + a + d = 180^\circ \)

\( \implies 3a = 180^\circ \)

\( \implies a = 60^\circ \)
Therefore, one angle is of \( 60^\circ \) and other is \( 90^\circ \) (given).
Let third angle be \( x^\circ \), then
\( 60^\circ + 90^\circ + x^\circ = 180^\circ \)

\( \implies 150^\circ + x^\circ = 180^\circ \)

\( \implies x^\circ = 180^\circ - 150^\circ = 30^\circ \)
Therefore, the angles of the right angled triangle are \( 30^\circ, 60^\circ, 90^\circ \).

Question. In an A.P., if \( S_n = 3n^2 + 2n \), then the value of ‘\( a_n \)’ is 
(a) 7n – 2
(b) 9n – 4
(c) 8n – 3
(d) 6n – 1
Answer: (d) 6n – 1
Explanation: Given: \( S_n = 3n^2 + 2n \)
\( S_1 = 3(1)^2 + 2 \times 1 = 3 + 2 = 5 \)

\( \implies a = 5 \)
\( S_2 = 3(2)^2 + 2 \times 2 = 3 \times 4 + 4 = 16 \)

\( \implies a_1 + a_2 = 16 \)

\( \implies a_1 = 5 \)

\( \implies a_2 = 11 \)
\( \therefore d = a_2 - a_1 = 11 - 5 = 6 \)
\( \therefore a_n = a + (n - 1)d \)
\( = 5 + (n - 1)6 = 5 + 6n - 6 = 6n - 1 \)

Question. The sum of (a + b), (a – b), (a – 3b), …….. to 22nd term is 
(a) 22a + 440b
(b) 22a – 440b
(c) 20a + 440b
(d) 22a – 400b
Answer: (b) 22a – 440b
Explanation: Given: \( a = a + b, d = a - b - a - b = -2b \)
\( \therefore S_{22} = \frac{22}{2} [2(a + b) + (22 - 1)(-2b)] \)
\( = 11 [2a + 2b + (21)(-2b)] \)

\( \implies S_{22} = 11 [2a + 2b - 42b] \)

\( \implies S_{22} = 11 [2a - 40b] \)
\( = 22a - 440b \)

Question. The first and last terms of an A.P. are 1 and 11. If their sum is 36, then the number of terms will be 
(a) 7
(b) 5
(c) 8
(d) 6
Answer: (d) 6
Explanation: Given: \( a = 1, l = 11 \) and \( S_n = 36 \)
\( \therefore S_n = \frac{n}{2} (a + l) \)

\( \implies 36 = \frac{n}{2} (1 + 11) \)

\( \implies 72 = n \times 12 \)

\( \implies n = 6 \)

Question. Is series \( \sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \dots \) an A.P.? Give reason. 
Answer: Common difference,
\( d_1 = \sqrt{6} - \sqrt{3} = \sqrt{3}(\sqrt{2} - 1) \)
\( d_2 = \sqrt{9} - \sqrt{6} = 3 - \sqrt{6} \)
\( d_3 = \sqrt{12} - \sqrt{9} = \sqrt{4 \times 3} - \sqrt{9} = 2\sqrt{3} - 3 \)
As common difference does not equal.
Hence, The given series is not in A.P.

Question. The sum of three numbers in AP is 21 and their product is 231. Find the numbers. 
Answer: Let the required numbers be (a-d), a and (a + d).............(1)
Then, according to question, (a - d) + a + (a + d) = 21

\( \implies 3a = 21 \)

\( \implies a = 7 \).
And, \( (a - d) \cdot a \cdot (a + d) = 231 \implies a(a^2 - d^2) = 231 \)

\( \implies 7(49 - d^2) = 231 \) [ \( \because a = 7 \) ]

\( \implies 7d^2 = 343 - 231 = 112 \)

\( \implies d^2 = 16 \)

\( \implies d = \pm 4 \).
Thus, \( a = 7 \) and \( d = \pm 4 \). Now substitute these values of a and d in above equation (1).
Therefore, the required numbers are (3,7,11) or (11,7,3).

Question. Find a and b such that the numbers a, 9, b, 25 form an AP. 
Answer: The numbers a, 9, b, 25 form an AP,
we have \( 9 - a = b - 9 = 25 - b \).
Now, \( b - 9 = 25 - b \implies 2b = 34 \implies b = 17 \).
And, \( 9 - a = b - 9 \implies a + b = 18 \implies a + 17 = 18 \implies a = 1 \).
Hence, \( a = 1 \) and \( b = 17 \).

Question. For an A.P., if \( a_{25} - a_{20} = 45 \), then find the value of d. 
Answer: Let the first term of an A.P be a and common difference d.
\( a_n = a + (n - 1)d \)
\( \therefore a_{25} - a_{20} = [a + (25 - 1)d] - [a + (20 - 1)d] \)
or, \( 45 = a + 24d - a - 19d \)
or, \( 45 = 5d \)
or, \( d = \frac{45}{5} = 9 \)

Question. Find the common difference of the AP : \( \frac{1}{p}, \frac{1-p}{p}, \frac{1-2p}{p}, \dots \) 
Answer: Common difference(d) = \( n^{th} term - (n - 1)^{th} term \)
\( \therefore d = a_2 - a_1 \)
\( d = \left( \frac{1-p}{p} \right) - \left( \frac{1}{p} \right) = \frac{(1-p)-(1)}{p} = \frac{-p}{p} = -1 \)
\( d = -1 \)

Question. An A.P. consists of 37 terms. The sum of the three middle most terms is 225 and the sum of the past three terms is 429. Find the A.P. 
Answer: Let the middle most terms of the A.P. be (a – d), a, (a + d)
Given \( (a - d) + a + (a + d) = 225 \)
\( 3a = 225 \)
or, \( a = 75 \)
and the middle term = \( \frac{37+1}{2} = 19^{th} \text{ term} \)
\( \therefore \) A.P. is (a - 18d ),....(a - 2d), {a - d), a, (a + d), (a + 2d),..........(a + 18d)
Sum of last three terms
\( (a + 18d) + (a + 17d) + (a + 16d) = 429 \)
or, \( 3a + 51d = 429 \)
or, \( 225 + 51d = 429 \)
or, \( 51d = 429 - 225 = 204 \)
or, \( d = 4 \)
First term \( a_1 = a - 18d = 75 - 18 \times 4 = 3 \)
\( a_2 = 3 + 4 = 7 \)
Hence, A.P. = 3, 7, 11 ,.........., 147

Question. Write the expression \( a_n - a_k \) for the AP: a, a + d, a + 2d, ... and find the common difference of the AP for which 20th term is 10 more than the 18th term. 
Answer: \( a_n = a + (n - 1)d; a_k = a + (k - 1)d \)
Now, \( a_n - a_k = [a + (n - 1)d] - [a + (k - 1)d] = (n - 1)d - (k - 1)d = (n - 1 - k + 1)d \)
\( a_n - a_k = (n - k)d \) .......(1)
Let \( a_{18} = x \).
\( a_{20} = x + 10 \)
Taking \( n = 20 \) and \( k = 18 \), equation (1) becomes
\( a_{20} - a_{18} = (20 - 18)d \)

\( \implies (x + 10) - x = 2d \)

\( \implies d = 5 \)

Question. The sum of the first three terms of an A.P. is 33. If the product of first and the third term exceeds the second term by 29, find the AP. 
Answer: Let the first three terms in A.P. be a - d, a, a + d. It is given that the sum of these terms is 33.
\( \therefore a - d + a + a + d = 33 \)

\( \implies 3a = 33 \)

\( \implies a = 11 \)
It is given that \( a_1 \times a_3 = a_2 + 29 \)
\( (a - d)(a + d) = a + 29 \)
\( a^2 - d^2 = a + 29 \)
\( 121 - d^2 = 11 + 29 \)
\( d^2 = 121 - 40 = 81 \)

\( \implies d = \pm 9 \)
If \( d = 9 \) then the series is 2, 11, 20, 29
If \( d = -9 \) then the series is 20, 11, 2, -7, -16

Question. If the \( m^{th} \) term of an AP be \( \frac{1}{n} \) and its \( n^{th} \) term be \( \frac{1}{m} \), then show that its (mn)th term is 1. 
Answer: Let a be the first term and d be the common difference of the given AP. Now, we know that in general \( m^{th} \) and \( n^{th} \) terms of the given A.P can be written as \( T_m = a + (m-1)d \) and \( T_n = a + (n-1)d \) respectively.
Now, \( T_m = \frac{1}{n} \) and \( T_n = \frac{1}{m} \) (given).
\( \therefore a + (m-1)d = \frac{1}{n} \) ......................(i)
and \( a + (n-1)d = \frac{1}{m} \) ...................... (ii)
On subtracting (ii) from (i), we get
\( (m-n)d = (\frac{1}{n} - \frac{1}{m}) = (\frac{m-n}{mn}) \)

\( \implies d = \frac{1}{mn} \)
Putting \( d = \frac{1}{mn} \) in (i), we get
\( a + \frac{(m-1)}{mn} \implies a = \{ \frac{1}{n} - \frac{(m-1)}{mn} \} = \frac{1}{mn} \)
Thus, \( a = \frac{1}{mn} \) and \( d = \frac{1}{mn} \)
\( \therefore \) Now, in general (mn)th term can be written as \( T_{mn} = a + (mn - 1)d \)
\( = \{ \frac{1}{mn} + \frac{(mn-1)}{mn} \} \) [ \( \because a = \frac{1}{mn} \) ]
= 1.
Hence, the (mn)th term of the given AP is 1.

Question. Find the sum of first 20 terms of an A.P., in which 3rd term is 7 and 7th term is two more than thrice of its 3rd term. 
Answer: Here, we have the sum of first 20 terms of an A.P., in which 3rd term is 7 and \( 7^{th} \) term is two more than thrice of its 3rd term.
Let "a" be the first term and" d"be the common difference of the given A.P. Therefore,
\( a_3 = 7 \) and \( a_7 = 3a_3 + 2 \) [Given]

\( \implies a + 2d = 7 \) and \( a + 6d = 3(a + 2d) + 2 \)

\( \implies a + 2d = 7 \) and \( a + 6d = 3a + 6d + 2 \)

\( \implies a + 2d = 7 \) and \( a - 3a = 6d - 6d + 2 \)

\( \implies a + 2d = 7 \) and \( -2a = 2 \)

\( \implies a + 2d = 7 \) and \( a = -1 \)

\( \implies -1 + 2d = 7 \)

\( \implies 2d = 7 + 1 = 8 \)

\( \implies d = 4 \)

\( \implies a = -1 \) and \( d = 4 \)
Putting \( n = 20, a = - 1 \) and \( d = 4 \) in \( S_n = \frac{n}{2} \{2a + (n - 1)d\} \), we get
\( S_{20} = \frac{20}{2} \{2 \times -1 + (20 - 1) \times 4\} = \frac{20}{2} (-2 + 76) = 740 \)

Question. The ratio of the sums of first m and first n terms of an A.P. is \( m^2 : n^2 \). Show that the ratio of its \( m^{th} \) and \( n^{th} \) terms is (2m - 1):(2n - 1 ). 
Answer: Let first term of given A.P. be a and common difference be d also sum of first m and first n terms be \( S_m \) and \( S_n \) respectively.
\( \therefore \frac{S_m}{S_n} = \frac{m^2}{n^2} \)
or, \( \frac{\frac{m}{2} [2a + (m-1)d]}{\frac{n}{2} [2a + (n-1)d]} = \frac{m^2}{n^2} \)
or, \( \frac{2a + (m-1)d}{2a + (n-1)d} = \frac{m^2}{n^2} \times \frac{n}{m} \)
or, \( \frac{2a + (m-1)d}{2a + (n-1)d} = \frac{m}{n} \)
or, \( m(2a + (n - 1)d) = n[2a + (m - 1)d] \)
Now, \( \frac{a_m}{a_n} = \frac{a + (m-1)d}{a + (n-1)d} \)
\( = \frac{a + (m-1) \times 2a}{a + (n-1) \times 2a} \)
or, \( = \frac{a + 2ma - 2a}{a + 2na - 2a} \)
or, \( = \frac{2ma - a}{2na - a} \)
or, \( = \frac{a(2m-1)}{a(2n-1)} \)
or, \( = \frac{(2m-1)}{(2n-1)} \)
= \( 2m - 1 : 2n - 1 \)
The ratio of its \( m^{th} \) and \( n^{th} \) terms is \( 2m - 1 : 2n - 1 \).
Hence proved

Question. In an A.P., the sum of first n terms is \( \frac{3n^2}{2} + \frac{13}{2}n \). Find its 25th term. 
Answer: According to the question,
Given Sum of n terms \( (S_n) = \frac{3n^2}{2} + \frac{13}{2}n \)
Put \( n = 24, S_{24} = \frac{3 \times 24 \times 24}{2} + \frac{13 \times 24}{2} \)
= 864 + 156
= 1020
Put \( n = 25, S_{25} = \frac{3 \times 25 \times 25}{2} + \frac{13 \times 25}{2} \)
= \( \frac{1875}{2} + \frac{325}{2} \)

\( \implies \frac{2200}{2} = 1100 \)
\( \therefore \) 25th term \( (a_{25}) = S_{25} - S_{24} \)
= 1100 - 1020
= 80

Question. Find the sum of all integers between 100 and 550 which are not divisible by 9. 
Answer: All integers between 100 and 550, which are divisible by 9
= 108, 117, 126,.........., 549
First term (a) = 108
Common difference(d) = 117 - 108 = 9
Last term(\( a_n \)) = 549

\( \implies a + (n - 1)d = 549 \)

\( \implies 108 + (n - 1)(9) = 549 \)

\( \implies 108 + 9n - 9 = 549 \)

\( \implies 9n = 549 + 9 - 108 \)

\( \implies 9n = 450 \)

\( \implies n = \frac{450}{9} = 50 \)
Sum of 50 terms = \( \frac{n}{2} [a + a_n] \)
\( = \frac{50}{2} [108 + 549] \)
= \( 25 \times 657 \)
= 16425
Now, sum of all integers between 100 and 550 which are not divisible by 9
= Sum of all integers between 100 and 550 - Sum of all integers between 100 and 550 which are divisible by 9
= [101 + 102 + 130 +......+ 549] - 16425
= \( \frac{549 \times 550}{2} - \frac{100 \times 101}{2} - 16425 \)
= 150975 - 5050 - 16425
= 129500

Question. If the sum of the first n terms of an A.P. is \( 4n - n^2 \), what is the first term? What is the sum of first two terms? What is the second term? Similarly, find the third, the tenth and the nth terms. 
Answer: Given that,
\( S_n = 4n - n^2 \)
First term, \( a = S_1 = 4(1) - (1)^2 = 4 - 1 = 3 \)
Sum of first two terms = \( S_2 \)
= \( 4(2) - (2)^2 = 8 - 4 = 4 \)
Second term, \( a_2 = S_2 - S_1 = 4 - 3 = 1 \)
\( d = a_2 - a = 1 - 3 = -2 \)
\( a_n = a + (n - 1)d \)
= \( 3 + (n - 1)(-2) \)
= \( 3 - 2n + 2 \)
= \( 5 - 2n \)
Therefore, \( a_3 = 5 - 2(3) = 5 - 6 = -1 \)
\( a_{10} = 5 - 2(10) = 5 - 20 = -15 \)
Hence, the sum of first two terms is 4. The second term is 1. \( 3^{rd}, 10^{th} \) and \( n^{th} \) terms are -1, -15, and \( 5 - 2n \) respectivey.

Question. Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
Answer: Sol. Let the first term be \( a \) and common difference be \( d \).
Now, we have
\( a_{11} = 38 \)
\( \implies \) \( a + (11 - 1)d = 38 \)
\( \implies \) \( a + 10d = 38 \) ...(i)
and \( a_{16} = 73 \)
\( \implies \) \( a + (16 - 1)d = 73 \)
\( \implies \) \( a + 15d = 73 \) ...(ii)
Now subtracting (ii) from (i), we have
Now, \( a + 10d = 38 \)
\( \underline{a + 15d = 73} \)
\( -5d = -35 \) or \( 5d = 35 \)
\( \therefore \) \( d = \frac{35}{5} = 7 \)
Putting the value of \( d \) in equation (i), we have
\( a + 10 \times 7 = 38 \)
\( \implies \) \( a + 70 = 38 \)
\( \implies \) \( a = 38 - 70 \)
\( \implies \) \( a = - 32 \)
We have, \( a = - 32 \) and \( d = 7 \)
Therefore, \( a_{31} = a + (31 - 1)d \)
\( \implies \) \( a_{31} = a + 30d \)
\( = (-32) + 30 \times 7 = - 32 + 210 \)
\( \implies \) \( a_{31} = 178 \)

Question. An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Answer: Sol. Let \( a \) be the first term and \( d \) be the common difference.
Since, given AP consists of 50 terms, so \( n = 50 \)
\( a_3 = 12 \)
\( \implies \) \( a + 2d = 12 \) ...(i)
Also, \( a_{50} = 106 \)
\( \implies \) \( a + 49d = 106 \) ...(ii)
Subtracting (i) from (ii), we have
\( 47d = 94 \)
\( \implies \) \( d = \frac{94}{47} = 2 \)
Putting the value of \( d \) in equation (i), we have
\( a + 2 \times 2 = 12 \)
\( \implies \) \( a = 12 - 4 = 8 \)
Here, \( a = 8, d = 2 \)
\( \therefore \) 29th term is given by
\( a_{29} = a + (29 - 1)d = 8 + 28 \times 2 \)
\( \implies \) \( a_{29} = 8 + 56 \)
\( \implies \) \( a_{29} = 64 \)

Question. How many three digit numbers are divisible by 7? 
Answer: Sol. Obviously first three digit number and last three digit number divisible by 7, are 105 and 994 respectively.
Now we have an AP
105, 112, 119, 126, ...... 994
Here, \( a = 105 \) and \( d = 7 \)
Let there be \( n \) terms in this AP.
\( \implies \) \( a_n = 994 \)
\( \implies \) \( a + (n - 1)d = 994 \)
\( \implies \) \( 105 + (n - 1)7 = 994 \)
\( \implies \) \( (n - 1)7 = 994 - 105 \)
\( \implies \) \( n - 1 = \frac{889}{7} \)
\( \implies \) \( n - 1 = 127 \)
\( \implies \) \( n = 127 + 1 = 128 \)
Hence 128 three digit numbers are divisible by 7.

Question. The sum of the 4th and 8th term of an AP is 24 and the sum of the 6th and 10th term is 44. Find the first three terms of the AP.
Answer: Sol. Let \( a \) be the first term and \( d \) the common difference of given AP.
Now, \( a_4 = a + (4 - 1)d = a + 3d \)
\( a_8 = a + (8 - 1)d = a + 7d \)
\( a_6 = a + (6 - 1)d = a + 5d \)
\( a_{10} = a + (10 - 1)d = a + 9d \)
From question,
\( a_4 + a_8 = 24 \)
\( \implies \) \( a + 3d + a + 7d = 24 \)
\( \implies \) \( 2a + 10d = 24 \)
\( \implies \) \( a + 5d = 12 \) ...(i)
Also, \( a_6 + a_{10} = 44 \)
\( a + 5d + a + 9d = 44 \)
\( \implies \) \( 2a + 14d = 44 \)
\( \implies \) \( a + 7d = 22 \) ...(ii)
Subtracting (i) from (ii), we get
\( a + 7d - a - 5d = 22 - 12 \)
\( \implies \) \( 2d = 10 \)
\( \implies \) \( d = 5 \)
Putting the value of \( d \) in (i), we get
\( a + 5 \times 5 = 12 \)
\( \implies \) \( a = 12 - 25 \)
\( \implies \) \( a = - 13 \)
Hence, first term = - 13.
Therefore, required three terms are
- 13, - 8, - 3.

Question. Subha Rao started work in 1995 at an annual salary of Rs. 5000 and received increment of Rs. 200 each year. In which year did her income reach Rs. 7000?
Answer: Sol. According to question, we have an AP.
Rs. 5000, Rs. 5200, Rs. 5400, ......
Obviously, 1st term \( a = \) Rs. 5000
Common difference \( d = \) Rs. 200
Let after \( n \) years it becomes Rs. 7000.
\( \implies \) \( a_n = \) Rs. 7000
\( \therefore \) \( a_n = a + (n - 1)d \)
\( \implies \) \( 7000 = 5000 + (n - 1) 200 \)
\( \implies \) \( 7000 - 5000 = (n - 1) 200 \)
\( \implies \) \( n - 1 = \frac{2000}{200} \)
\( \implies \) \( n - 1 = 10 \)
\( \implies \) \( n = 11 \)
Hence in 11th year her income will reach Rs. 7000.

Question. How many terms of the AP 9, 17, 25, ... must be taken to give a sum of 636? 
Answer: Sol. \( a = 9 \), \( d = 8 \), \( S_n = 636 \).
\( S_n = \frac{n}{2} [ 2a + (n-1)d ] \)
\( 636 = \frac{n}{2} [ 18 + (n-1)8 ] \)
\( 636 = n [ 9 + (n-1)4 ] \)
\( 636 = n ( 9 + 4n - 4 ) \)
\( 636 = n ( 5 + 4n ) \)
\( 636 = 5n + 4n^2 \)
\( 4n^2 + 5n - 636 = 0 \)
\( 4n^2 + 53n - 48n - 636 = 0 \)
\( n(4n + 53) - 12(4n + 53) = 0 \)
\( (4n + 53)(n - 12) = 0 \)
\( \therefore \) \( n = \frac{-53}{4} \) or \( 12 \).
as \( n \) is a natural number, \( n = 12 \).
\( \therefore \) 12 terms are required to give sum 636.

Question. Find the sum of the first 15 multiples of 8. 
Answer: Sol. The first 15 multiples of 8 are
8, 16, 24, ... 120
Clearly, these numbers are in AP 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. A sum of Rs. 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs. 20 less than its preceding prize, find the value of each of the prizes.
Answer: Sol. Let the prizes be \( a + 60, a + 40, a + 20, a, a - 20, a - 40, a - 60 \).
Therefore, the sum of prizes is
\( a + 60 + a + 40 + a + 20 + a + a - 20 + a - 40 + a - 60 = 700 \)
\( \implies \) \( 7a = 700 \)
\( \implies \) \( a = \frac{700}{7} = 100 \)
Thus, the value of seven prizes are
100 + 60, 100 + 40, 100 + 20, 100, 100 - 20, 100 - 40, 100 - 60
i.e., Rs. 160, Rs. 140, Rs. 120, Rs. 100, Rs. 80, Rs. 60, Rs. 40

Question. 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see figure). In how many rows are the 200 logs placed and how many logs are in the top row? 
Answer: Sol. Since, logs are stacked in each row form a series 20 + 19 + 18 + 17 + .... Clearly, it is an AP with first term, \( a = 20 \) and common difference, \( d = 19 - 20 = - 1 \).
\( S_n = 200 \)
\( S_n = \frac{n}{2} [2a + (n-1)d] \)
\( 200 = \frac{n}{2} [2 \times 20 + (n-1)(-1)] \)
\( \implies \) \( 400 = n(40 - n + 1) \)
\( \implies \) \( n^2 - 41n + 400 = 0 \)
\( n^2 - 25n - 16n + 400 = 0 \) (By factorisation)
\( \implies \) \( n(n - 25) - 16(n - 25) = 0 \)
\( \implies \) \( (n - 25)(n - 16) = 0 \)
\( \implies \) \( n = 16 \) and \( n = 25 \)
Hence, the number of rows is either 25 or 16.
If \( n = 16 \) then \( l_n = a + (n - 1)d \)
\( = 20 + (16 - 1)(-1) \)
\( = 20 - 15 = 5 \)
If \( n = 25 \) then \( l_n = a + (n - 1)d \)
\( = 20 + (25 - 1)(-1) \)
\( = 20 - 24 = - 4 \) (Not possible)
Hence, the number of row is 16 and number of logs in the top row = 5.

Question. Which term of the AP 121, 117, 113, ... is its first negative term?
Answer: Sol. Given, first term, \( a = 121 \), common difference, \( d = 117 - 121 = - 4 \)
\(\because\) \( n \)th term of an AP,
\( a_n = a + (n - 1)d \)
\( = 121 + (n - 1) \times (- 4) \)
\( = 121 - 4n + 4 = 125 - 4n \)
For first negative term, \( a_n < 0 \).
\( \implies \) \( 125 - 4n < 0 \)
\( \implies \) \( 125 < 4n \)
\( \implies \) \( 4n > 125 \)
\( \implies \) \( n > \frac{125}{4} \)
\( \implies \) \( n > 31\frac{1}{4} \)
Least integral value of \( n = 32 \).
Hence, 32nd term of the given AP is the first negative term.

Very Short Answer Questions

Question. For what value of \( k \) will \( k + 9 \), \( 2k - 1 \) and \( 2k + 7 \) be the consecutive terms of an AP? 
Answer: Sol. We have-
Three consecutive terms of AP = \( k+9 \), \( 2k-1 \), \( 2k+7 \)
Then,
\( (k + 9) + (2k + 7) = 2(2k - 1) \) \(\{a + c = 2b\}\)
\( \implies \) \( k + 9 + 2k + 7 = 4k - 2 \)
\( 3k + 16 = 4k - 2 \)
\( 16 + 2 = 4k - 3k \)
\( 18 = k \)

Question. Write the common difference of the AP \( \sqrt{3}, \sqrt{12}, \sqrt{27}, \sqrt{48}, ... \) 
Answer: Sol. Given AP, \( \sqrt{3}, \sqrt{12}, \sqrt{27}, \sqrt{48}, ... \)
\( \implies \) \( \sqrt{3}, 2\sqrt{3}, 3\sqrt{3}, 4\sqrt{3}, ... \)
\( \therefore \) Common difference = \( 2\sqrt{3} - \sqrt{3} = \sqrt{3} \)
\( \implies \) \( d = \sqrt{3} \)

Question. In an AP, if the common difference(d) = - 4, and the seventh term (\( a_7 \)) is 4, then find the first term. 
Answer: Sol. \( d = -4, a_7 = 4 \).
\( t_n = a + (n-1)d \)
\( a_7 = a + (7-1)(-4) \)
\( 4 = a + 6 \times (-4) \)
\( a = 24 + 4 \)
\( a = 28 \).
The first term is 28.

Question. If 7 times the \( 7^{th} \) term of an AP is equal to 11 times its \( 11^{th} \) term, then find its \( 18^{th} \) term. 
Answer: Sol. Given, \( 7a_7 = 11a_{11} \)
\( \implies \) \( 7(a + 6d) = 11(a + 10d) \) or \( 7a + 42d = 11a + 110d \)
\( \implies \) \( 4a + 68d = 0 \) or \( a + 17d = 0 \)
Now, \( a_{18} = a + 17d = 0 \)

Question. Find the number of terms in the AP: 18, 15 1/2, 13, ..., - 47. 
Answer: Sol. Here \( -47 = 18 + (n - 1) \left( - \frac{5}{2} \right) \). Here, \( d = - \frac{5}{2} \)
\( \implies \) \( n = 27 \)
[CBSE Marking Scheme 2019(30/4/2)]

Question. Find the 9th term from the end (towards the first term) of the AP 5, 9, 13, ..., 185. 
Answer: Sol. \( l = 185, d = 4 \)
\( l_9 = l - (n - 1)d = 185 - 8 \times 4 = 153 \)

Question. Find the sum of first 10 multiples of 6. 
Answer: Sol. First 10 multiples of 6 form AP \( \implies \) 6, 12, 18 ... 60.
Sum of 1st 10 multiples = \( \frac{n}{2} [a + l] \)
\( = \frac{10}{2} [6 + 60] \)
\( = 5 \times 66 \)
\( = 330 \)

Question. Find the sum of the first 100 natural numbers. 
Answer: Sol. We have, Sum of first \( n \) natural numbers = \( \frac{n(n + 1)}{2} \)
\( \therefore \) Sum of first 100 natural numbers = \( \frac{100(100 + 1)}{2} = \frac{100 \times 101}{2} = 50 \times 101 = 5050 \)

Short Answer Questions-I

Question. Find the middle term of the AP 213, 205, 197, ....., 37. 
Answer: Sol. We have, \( a = 213, d = 205 - 213 = - 8 \) and \( l = 37 \)
Let \( n \) be the number of terms of the AP.
\( \therefore \) \( l = a + (n - 1)d \)
\( \implies \) \( 37 = 213 + (n - 1) \times (-8) \)
\( \implies \) \( (n - 1) = \frac{-176}{-8} = 22 \)
\( \implies \) \( n = 23 \)
\( \therefore \) The middle term will be \( \left( \frac{n+1}{2} \right)^{th} \) term
i.e., \( \left( \frac{23+1}{2} \right)^{th} = 12^{th} \) term
\( a_{12} = a + 11d = 213 + 11(-8) = 125 \)

Question. For what value of \( n \), are the \( n \)th terms of two APs 63, 65, 67,... and 3, 10, 17,... equal? 
Answer: Sol. Let \( n \)th terms for two given series be \( a_n \) and \( a'_n \).
According to question,
\( a_n = a'_n \)
\( \implies \) \( a + (n - 1)d = a' + (n - 1)d' \)
\( \implies \) \( 63 + (n - 1)2 = 3 + (n - 1)7 \)
\( \implies \) \( 5n = 65 \)
\( \implies \) \( n = 13 \).

Question. If the ratio of sum of the first \( m \) and \( n \) terms of an AP is \( m^2 : n^2 \), show that the ratio of its \( m \)th and \( n \)th terms is \( (2m - 1) : (2n - 1) \). 
Answer: Sol. \( \frac{S_m}{S_n} = \frac{m^2}{n^2} \)
\( \implies \) \( \frac{\frac{m}{2} [2a + (m-1)d]}{\frac{n}{2} [2a + (n-1)d]} = \frac{m^2}{n^2} \)
\( \implies \) \( \frac{m}{n} \frac{2a + (m-1)d}{2a + (n-1)d} = \frac{m^2}{n^2} \)
\( \implies \) \( 2am + mnd - md = 2an + mnd - nd \)
\( \implies \) \( a(2m - 2n) = d(m - n) \)
\( \implies \) \( 2a = d \)
\( \frac{a_m}{a_n} = \frac{a + (m-1)d}{a + (n-1)d} = \frac{a + 2(m-1)a}{a + 2(n-1)a} = \frac{2m-1}{2n-1} \). Hence proved.

Question. Determine the AP whose third term is 16 and \( 7^{th} \) term exceeds the 5th term by 12. 
Answer: Sol. Here \( t_3 = 16 \) and \( t_7 = t_5 + 12 \)
\( \implies \) \( a + 2d = 16 \) ...(i) and \( a + 6d = a + 4d + 12 \) ...(ii)
From (ii), \( d = 6 \)
From (i), \( a = 4 \)
\( \therefore \) AP is 4, 10, 16, ...
[CBSE Marking Scheme 2019(30/4/2)]

Question. Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5. 
Answer: Sol. Natural numbers between 101 and 999 divisible by both 2 and 5 are 110, 120, ... 990.
So, \( a = 110, d = 10, a_n = 990 \)
We know, \( a_n = a + (n - 1)d \)
\( 990 = 110 + (n - 1) 10 \)
\( (n - 1) = \frac{990 - 110}{10} \)
\( \implies \) \( n = 88 + 1 = 89 \)

Question. If the \( 17^{th} \) term of an AP exceeds its \( 10^{th} \) term by 7, find the common difference. 
Answer: Sol. According to question,
\( a_{17} = a_{10} + 7 \)
\( \implies \) \( a + (17 - 1) d = a + (10 - 1) d + 7 \)
\( \implies \) \( a + 16d = a + 9d + 7 \)
\( \implies \) \( 16d - 9d = 7 \)
\( \implies \) \( 7d = 7 \)
\( \implies \) \( d = 1 \)
Common difference = 1

Question. Solve the equation: 1 + 5 + 9 + 13 + ..... + \( x \) = 1326. 
Answer: Sol. Given, 1 + 5 + 9 + 13 + ... + \( x \) = 1326
It is an AP with \( a = 1, d = 4, a_n = x = l \) and \( S_n = 1326 \)
\( a_n = a + (n - 1)d \)
\( x = 1 + (n - 1) \times 4 \)
\( \implies \) \( \frac{(x - 1)}{4} = n - 1 \)
\( \implies \) \( n = \frac{x - 1}{4} + 1 \)
\( n = \frac{x + 3}{4} \)
Now, \( S_n = \frac{n}{2} (a + l) \)
\( 1326 = \frac{\frac{x + 3}{4}}{2} (1 + x) \)
\( \implies \) \( \frac{(x + 1)(x + 3)}{8} = 1326 \)
\( \implies \) \( x^2 + 4x + 3 = 10608 \)
\( \implies \) \( x^2 + 4x - 10605 = 0 \)
\( x = \frac{-4 \pm \sqrt{16 + 42420}}{2} \)
\( \implies \) \( x = \frac{-4 \pm \sqrt{42436}}{2} = \frac{-4 \pm 206}{2} \)
\( x = \frac{-4 + 206}{2} = \frac{202}{2} = 101 \) [Ignoring negative value]
\( x = 101 \)

Question. The first and the last term of an AP are 5 and 45 respectively. If the sum of all its terms is 400, find its common difference. 
Answer: Sol. Let the first term be '\( a \)' and common difference be '\( d \)'.
Given, \( a = 5, T_n = 45, S_n = 400 \)
\( T_n = a + (n - 1)d \)
\( \implies \) \( 45 = 5 + (n - 1)d \)
\( \implies \) \( (n - 1)d = 40 \) ...(i)
\( S_n = \frac{n}{2} (a + T_n) \)
\( \implies \) \( 400 = \frac{n}{2} (5 + 45) \)
\( \implies \) \( n = 2 \times 8 = 16 \)
Substituting the value of \( n \) in (i)
\( (16 - 1)d = 40 \)
\( \implies \) \( d = \frac{40}{15} = \frac{8}{3} \)

Question. If \( S_n \), the sum of the first \( n \) terms of an AP is given by \( S_n = 2n^2 + n \), then find its \( n \)th term. 
Answer: Sol. We have in an AP
\( S_n = 2n^2 + n \)
\( \therefore \) \( S_{n-1} = 2(n - 1)^2 + (n - 1) = 2n^2 - 4n + 2 + n - 1 \)
\( \implies \) \( S_{n-1} = 2n^2 - 3n + 1 \)
\( \therefore \) Its \( n \)th term, \( t_n = S_n - S_{n-1} = 2n^2 + n - 2n^2 + 3n - 1 \)
\( t_n = 4n - 1 \)