CBSE Class 10 Maths HOTs Arithmetic Progressions Set 13

Very Short Questions

Question. Find the sum of the first 100 natural numbers. 
Answer: The list of first 100 natural numbers. 1, 2, 3, ......, 100, which forms an AP with \( a = 1, d = 1 \)
So, \( S_{100} = \frac{100}{2} [2 \times 1 + (100 - 1)(1)] \)
\( = 50 [101] \)
\( = 5050 \)

Question. If the mean of the first \( n \) natural number is 15, then find \( n \). 
Answer: The natural number are, 1, 2, 3 ..... \( n \)
Their, mean \( = \frac{S_n}{n} \) and \( S_n = \frac{n}{2} [2a + (n - 1)d] \)
The mean of first \( n \) natural number is \( \frac{n(n + 1)}{2n} \) is \( \frac{n + 1}{2} \)
\( \therefore \frac{n + 1}{2} = 15 \), (given)
\( \implies n = 29 \)

Question. If in an A.P., \( a = 15, d = -3 \) and \( a_n = 0 \), then find the value of \( n \). 
Answer: Given, for an A.P., \( a = 15, d = -3 \) and \( a_n = 0 \)
Now, \( a_n = a + (n - 1)d \)
\( \implies 0 = 15 + (n - 1) \times (-3) \)
\( \implies (n - 1) = 5 \)
\( \implies n = 6 \)
Hence, the value of \( n \) is 6.

Question. Find the number of terms in the A.P.: 18, \( 15\frac{1}{2} \), 13, ..., -47. 
Answer: Given A.P. is 18, \( 15\frac{1}{2} \), 13, ....., -47
Here, first term, \( a = 18 \)
Common difference, \( d = \frac{31}{2} - 18 = \frac{31 - 36}{2} = \frac{-5}{2} \)
last term, \( a_n = -47 \)
Now, \( a_n = a + (n - 1)d \), where, 'n' is the number of terms.
\( -47 = 18 + (n - 1) \times \frac{-5}{2} \)
\( \implies (n - 1) \times \frac{-5}{2} = -65 \)
\( \implies (n - 1) = 26 \implies n = 27 \)
Hence, the number of terms in the given A.P. is 27.

Question. Find the common difference of the Arithmetic Progression (A.P.) 
\( \frac{1}{a}, \frac{3-a}{3a}, \frac{3-2a}{3a}, \dots (a \neq 0) \)
Answer: Given: Arithmetic progression (AP) is \( \frac{1}{a}, \frac{3-a}{3a}, \frac{3-2a}{3a}, \dots (a \neq 0) \)
In the given progression,
\( a_1 = \frac{1}{a}, a_2 = \frac{3-a}{3a}, a_3 = \frac{3-2a}{3a} \)
Common difference,
\( d = a_2 - a_1 = a_3 - a_2 \)
\( = \frac{3-a}{3a} - \frac{1}{a} \) or \( \frac{3-2a}{3a} - \frac{3-a}{3a} \)
\( = \frac{3-a-3}{3a} \) or \( \frac{3-2a-3+a}{3a} \)
\( = \frac{-a}{3a} \) or \( \frac{-a}{3a} \)
\( = \frac{-1}{3} \)
Hence, the common difference of the A.P. is \( \frac{-1}{3} \).

Question. Justify whether it is true to say that -1, \( -\frac{3}{2} \), -2, \( \frac{5}{2} \), ... form an AP as \( a_2 - a_1 = a_3 - a_2 \).
Answer: False
Explanation: The given series of numbers is -1, \( -\frac{3}{2} \), -2, \( \frac{5}{2} \), ...
Here, \( a_1 = -1, a_2 = -\frac{3}{2}, a_3 = -2, a_4 = \frac{5}{2}, \dots \)
Difference between two successive terms
\( a_2 - a_1 = -\frac{3}{2} - (-1) = -\frac{3}{2} + 1 = \frac{-3+2}{2} = \frac{-1}{2} \)
\( a_3 - a_2 = -2 - (-\frac{3}{2}) = -2 + \frac{3}{2} = \frac{-4+3}{2} = \frac{-1}{2} \)
\( a_4 - a_3 = \frac{5}{2} - (-2) = \frac{5}{2} + 2 = \frac{5+4}{2} = \frac{9}{2} \)
We can see that \( a_2 - a_1 = -\frac{1}{2}, a_3 - a_2 = -\frac{1}{2} \),
\( a_4 - a_3 = \frac{9}{2} \).
\( \implies a_2 - a_1 = a_3 - a_2 \)
but \( a_3 - a_2 \neq a_4 - a_3 \)
Clearly, the difference of two successive terms is not the same, hence we can say that the given list of numbers does not form an AP.

Question. How many 2-digit numbers are divisible by 3? 
Answer: Two digits numbers divisible by 3 are 12, 15, 18 ..., 99
Here, first term, \( a = 12 \)
common difference, \( d = 15 - 12 = 18 - 15 = 3 \)
last term, \( a_n = 99 \)
Now, nth term, \( a_n = a + (n - 1)d \),
where, 'n' is the number of the terms
\( \implies 99 = 12 + (n - 1) 3 \)
\( \implies 99 = 12 + 3n - 3 \)
\( \implies 3n = 90 \)
\( \implies n = 30 \)
Hence, the number of terms are 30.

Question. In an AP, if the common difference \( d = -4 \), and the seventh term (\( a_7 \)) is 4, then find the first term.
Answer: Given common difference \( d \) of an A.P. \( = -4 \)
Seventh term (\( a_7 \)) of A.P. \( = 4 \)
Let, 'a' be the first term of the A.P.
Then, \( a_7 = a + (n - 1)d \)
\( \implies 4 = a + (7 - 1) (-4) \)
\( \implies 4 = a - 24 \)
\( \implies a = 28 \)
Hence, the first term of the A.P. is 28.

Question. Write the nth term of the A.P. \( \frac{1}{m}, \frac{1+m}{m}, \frac{1+2m}{m}, \dots \)
Answer: Given, A.P. is \( \frac{1}{m}, \frac{1+m}{m}, \frac{1+2m}{m}, \dots \)
Here, first term \( a = \frac{1}{m} \)
Common difference, \( d = \frac{1+m}{m} - \frac{1}{m} = \frac{m}{m} = 1 \)
\( \therefore \) nth term, \( a_n = a + (n - 1)d \)
\( = \frac{1}{m} + (n - 1) \times 1 \)
\( = \frac{1}{m} + n - 1 \)
\( = \frac{1 + mn - m}{m} \)
\( = \frac{1 + m(n - 1)}{m} \)
Hence, the \( n^{th} \) term of given A.P. is \( \frac{mn - m + 1}{m} \).

Question. If the nth term of the A.P. -1, 4, 9, 14, .... is 129, find the value of \( n \). 
Answer: Given, A.P. is -1, 4, 9, 14, ......
nth term, \( a_n = 129 \)
Here, first term, \( a = -1 \)
Common difference, \( d = 4 - (-1) = 5 \)
\( \therefore a + (n - 1)d = 129 \)
\( \implies -1 + (n - 1)5 = 129 \)
\( \implies -1 + 5n - 5 = 129 \)
\( \implies 5n - 6 = 129 \)
\( \implies 5n = 135 \)
\( \implies n = 27 \)
Hence, the value of \( n \) is 27.

Question. Find the 9th term from the end (towards the first term) of the A.P. 5, 9, 13, ...., 185.
Answer: Given A.P. is 5, 9, 13, ...., 185. Since we have to find the 9th term from the end,
We should reverse the A.P. Then, it will become easier to find the 9th term from the starting:
185, 181, ......, 13, 9, 5
Now, first term, \( a = 185 \)
Common difference, \( d = 181 - 185 = -4 \)
Then, 9th term of A.P.,
\( a_9 = a + (9 - 1) \times d \)
[\( \because a_n = a + (n - 1)d \)]
\( = 185 + 8 \times (-4) \)
\( = 185 - 32 = 153 \)
Hence, the 9th term from the end is 153.

Question. For the AP: -3, -7, -11, ..., can we directly find \( a_{30} - a_{20} \) without actually finding \( a_{30} \) and \( a_{20} \)? Give reasons for your answer.
Answer: Yes, we can find.
The given list of numbers of an AP -3, -7, -11 ...
We know that \( a_n = a + (n - 1)d \)
\( \implies a_{30} = a + (30 - 1)d = a + 29d \)
\( \implies a_{20} = a + (20 - 1)d = a + 19d \)
Now, \( a_{30} - a_{20} = (a + 29d) - (a + 19d) \)
\( = a + 29d - a - 19d = 10d \) ...(i)
For the given series, common difference,
\( d = -7 - (-3) = -7 + 3 = -4 \)
Hence, \( a_{30} - a_{20} = 10d = 10(-4) = -40 \) [Using eqn. (i)]
\( \implies a_{30} - a_{20} = -40 \)

Question. If the first three terms of an A.P. are \( b \), \( c \) and \( 2b \), then find the ratio of \( b \) and \( c \).
Answer: \( b, c \) and \( 2b \) are in A.P
\( \implies c = \frac{3b}{2} \)
\( \therefore b : c = 2 : 3 \)

Question. Find the 16th term of the AP: 2, 7, 12, 17, ...
Answer: Here, first term, \( a = 2 \) and common difference, \( d = 5 \).
Using formula, nth term,
\( a_n = a + (n - 1)d \)
So, \( a_{16} = a + 15d \)
\( = 2 + 15 \times 5 \)
\( = 2 + 75 = 77 \)

Question. Find the mean of first eleven natural numbers.
Answer: Mean \( = \frac{1 + 2 + 3 + ... + 11}{11} = \frac{11(11 + 1)}{2 \times 11} = 6 \)
\( \therefore \) Sum of first n numbers \( \left[ \frac{n(n + 1)}{2} \right] \)
 

Short Answer (SA-I) Type Questions

Question. Show that \( (a – b)^2 \), \( (a^2 + b^2) \) and \( (a + b)^2 \) are in AP.
Answer: \( (a – b)^2 \), \( (a^2 + b^2) \) and \( (a + b)^2 \) will be in AP, if \( 2(a^2 + b^2) = (a – b)^2 + (a + b)^2 \)
L.H.S. \( = 2(a^2 + b^2) \)
R.H.S. \( = (a – b)^2 + (a + b)^2 \)
\( = a^2 + b^2 – 2ab + a^2 + b^2 + 2ab \)
\( = 2a^2 + 2b^2 = 2(a^2 + b^2) \)
\( \therefore L.H.S. = R.H.S. \)
Hence, \( (a – b)^2 \), \( (a^2 + b^2) \) and \( (a + b)^2 \) are in A.P.

Question. If the 17th term of an A.P. exceeds its 10th term by 7, find the common difference. 
Answer: Given : \( a_{17} = a_{10} + 7 \) ...(i)
Let ‘a’ be the first term of A.P. and ‘d’ be its common difference.
Then \( a_{10} = a + (10 – 1)d = a + 9d \)
and \( a_{17} = a + (17 – 1)d = a + 16d \)
put the values in (i), we get
\( a + 16d = a + 9d + 7 \)
\( \implies 7d = 7 \)
\( \implies d = 1 \)
Hence, the common difference of given A.P. is 1.

Question. How many multiples of 4 lie between 10 and 205 ?
Answer: Multiples of 4 between 10 and 205 is 12, 16, 20, 24, 28, ......, 204.
Let, the number of multiples be ‘n’.
Here, first term, \( a = 12 \)
common difference, \( d = 4 \)
last term, \( a_n = 204 \)
Since, \( n^{th} \) term of an A.P. is \( a_n = a + (n – 1)d \)
\( \implies 204 = 12 + (n – 1) \times 4 \)
\( \implies 204 = 12 + 4n – 4 \)
\( \implies 4n = 204 – 8 = 196 \)
\( \implies n = \frac{196}{4} = 49 \)
Hence, the number of multiples of 4 are 49.

Question. Determine the A.P. whose third term is 16 and 7th term exceeds the 5th term by 12.
Answer: Let, the first term of an A.P. be ‘a’ and common difference be ‘d’.
Given, \( a_3 = 16 \)
i.e., \( a + (3 – 1)d = 16 \)
\( \implies a + 2d = 16 \) ...(i)
and \( a_7 = a_5 + 12 \) (given)
\( \implies a + (7 – 1)d = a + (5 – 1)d + 12 \)
\( \implies a + 6d = a + 4d + 12 \)
\( \implies 2d = 12 \)
\( \implies d = 6 \)
If we put the value of d in equation (i), we get
\( a + 2 \times 6 = 16 \)
\( \implies a = 4 \)
\( \therefore \) The first term of the A.P is 4 and its common difference is 6.
Hence, the required A.P. is 4, 10, 16, 20........

Question. Two APs have the same common difference. The first term of one AP is 2 and that of the other is 7. The difference between their 10th terms is the same as the difference between their 21st terms, which is the same as the difference between any two corresponding terms. Why? 
Answer: Let d be the common difference of two AP's
First term of first AP is 2.
First AP will be 2, 2 + d, 2 + 2d ...
First term of second AP is 7 second AP will be 7, 7 + d, 7 + 2d ...
We know that \( a_n = a + (n – 1)d \)
10th term of first AP = 2 + 9d
10th term of second AP = 7 + 9d
Difference of their 10th term = \( (7 + 9d) – (2 + 9d) \)
\( = 7 + 9d – 2 – 9d = 5 \)
Again, 21st term of first AP = 2 + 20d
21st term of second AP = 7 + 20d
Difference of their 21st term \( = (7 + 20d) – (2 + 20d) \)
\( = 7 + 20d – 2 – 20d = 5 \)
Thus, we can say that if \( a_n \) and \( b_n \) are \( n^{th} \) terms of first and second AP respectively, then
\( b_n – a_n = [7 + (n – 1)d] – [2 + (n – 1)d] \)
\( = 7 + (n – 1)d – 2 – (n – 1)d = 5 \)
\( \implies b_n – a_n = 5 \)
Hence, the difference between any two corresponding terms of such AP’s is the same as the difference between their first terms.

Question. Which term of the AP 3, 15, 27, 39, .... will be 120 more than its 21st term?
Answer: Given: A.P. is 3, 15, 27, 39, .......
Here, the first term, \( a = 3 \)
Common difference, \( d = 15 – 3 = 27 – 15 = 12 \).
Now, 21st term \( a_{21} = 3 + (21 – 1) \times 12 \)
\( = 3 + 20 \times 12 = 3 + 240 = 243 \)
According to the given condition :
\( a_n = a_{21} + 120 \)
where, \( a_n \) is the \( n^{th} \) term
\( a_n = 243 + 120 = 363 \)
\( \implies a + (n – 1)d = 363 \)
\( \implies 3 + (n – 1) \times 12 = 363 \)
\( \implies (n – 1) = \frac{360}{12} \)
\( \implies n = 31 \)
Hence, the 31st term is 120 more than the 21st term.

Question. If \( S_n \), the sum of first n terms of an AP is given by \( S_n = 3n^2 – 4n \), find the nth term.
Answer: Here, the sum of first n terms \( S_n = 3n^2 – 4n \)
\( \therefore S_{n – 1} = 3 (n – 1)^2 – 4(n – 1) \)
\( = 3n^2 – 10n + 7 \)
Hence, the nth term, \( a_n = S_n – S_{n – 1} \)
\( = (3n^2 – 4n) – (3n^2 – 10n + 7) \)
\( = 3n^2 – 4n – 3n^2 + 10n – 7 \)
\( a_n = 6n – 7 \)
Hence, the nth term is 6n – 7.

Question. Find the sum of first 8 multiples of 3.
Answer: First 8 multiple of 3 are: 3, 6, 9, 12, 15, 18, 21, 24, which forms an A.P.
\( S_8 = 3 + 6 + 9 + 12 + ...... + 24 \)
Here, first term, \( a = 3 \)
common difference, \( d = 3 \)
number of terms, \( n = 8 \)
\( S_n = \frac{n}{2} [2a + (n – 1)d] \)
\( = \frac{8}{2} [2 \times 3 + (8 – 1)3] \)
\( = 4[6 + 21] = 4 \times 27 = 108 \)
Thus, the sum of the first 8 multiples is 108.

Question. If seven times the 7th term of an A.P. is equal to eleven times the 11th term, then what will be its 18th term?
Answer: Let the first term of the A.P. be ‘a’ and its common difference be ‘d’.
Given, \( 7a_7 = 11a_{11} \)
Then, \( 7(a + 6d) = 11(a + 10d) \)
[ \( \because a_n = a + (n – 1)d \) ]
\( \implies 7a + 42d = 11a + 110d \)
\( \implies 7a – 11a = 110d – 42d \)
\( \implies – 4a = 68d \)
\( \implies a = – 17d \)
Now, 18th term of A.P.
\( a_{18} = a + (18 – 1)d \)
\( = a + 17d \)
putting \( a = – 17d \)
\( = –17d + 17d \)
\( a_{18} = 0 \)
Hence, the 18th term of A.P. is 0.

Question. The 10th term of an A.P. is –4 and its 22nd term is (– 16). Find its 38th term.
Answer: Let, the first term of A.P. be ‘a’ and its common difference be ‘d’.
\( a_{10} = – 4 \) (given)
i.e. \( a + 9d = – 4 \) ...(i)
\( a_{22} = – 16 \) (given)
\( a + 21d = – 16 \) ...(ii)
On solving (i) and (ii), we get
\( a + 9d = – 4 \)
\( a + 21d = – 16 \)
----------------
\( – 12d = 12 \)
\( d = – 1 \)
If we put the value of \( d = – 1 \) in equation (i), we get
\( a = – 4 + 9 = 5 \)
\( \therefore \) first term, \( a = 5 \)
and common difference, \( d = – 1 \)
38th term, \( a_{38} = a + (38 – 1)d \)
\( = 5 + 37(– 1) = – 32 \)
Hence, the 38th term of the A.P. is – 32.

Question. Find how many integers between 200 and 500 are divisible by 8.
Answer: Integers between 200 and 500 divisible by 8 are 208, 216, 224 ,..., 496.
This series forms an A.P., where first term, \( a = 208 \)
common difference, \( d = 8 \)
and last term, \( l = 496 \)
Let, the number of integers be ‘n’
Then, \( n^{th} \) term = \( a_n \) (last term l) \( = a + (n – 1)d \)
\( 496 = 208 + (n – 1) \times 8 \)
\( \implies (n – 1) \times 8 = 496 – 208 = 288 \)
\( \implies (n – 1) = \frac{288}{8} = 36 \)
\( \implies n = 37 \)
Hence, the number of terms are 37.

Question. Determine the AP whose third term is 5 and the seventh term is 9.
Answer: Given, third term of A.P., \( a_3 = 5 \)
Seventh term of A.P., \( a_7 = 9 \)
Let the first term of A.P. be ‘a’ and its common difference be ‘d’.
Now, \( a_3 = a + (3 – 1)d \)
or \( 5 = a + 2d \) ...(i)
and \( a_7 = a + (7 – 1)d \)
or \( 9 = a + 6d \) ...(ii)
On solving (i) and (ii), we get
\( 4d = 4 \)
\( \implies d = 1 \)
If we put the value of ‘d’ in equation (i), we get
\( 5 = a + 2 \times 1 \)
\( \implies a = 3 \)
Hence, the A.P. is 3, 4, 5, 6.

Question. If the sum of the first 9 terms of an AP is equal to the sum of its first 11 terms, then find the sum of its first 20 terms.
Answer: Let, the first term of an A.P. be ‘a’ and its common difference be ‘d’.
Given, \( S_9 = S_{11} \)
\( \implies \frac{9}{2} [2a + (9 – 1)d] = \frac{11}{2} [2a + (11 – 1)d] \)
[ \( \because S_n = \frac{n}{2} [2a + (n – 1)d] \) ]
\( \implies 9[2a + 8d] = 11[2a + 10d] \)
\( \implies 18a + 72d = 22a + 110d \)
\( \implies 4a = – 38d \)
\( \implies 2a = – 19d \) ...(i)
Now, sum of first 20 terms:
\( S_{20} = \frac{20}{2} [2a + (20 – 1)d] \)
\( = 10[2a + 19d] \)
\( = 10[– 19d + 19d] \) [from (i)]
\( = 10 \times 0 = 0 \)
Hence, the sum of first 20 terms is 0.

Question. Find the number of natural numbers between 102 and 998 which are divisible by 2 and 5 both. 
Answer: 110, 120, 130, … , 990
\( a_n = 990 \)
\( \implies 110 + (n - 1) \times 10 = 990 \)
\( \therefore n = 89 \)

Question. The common difference between the terms of two APs is same. If the difference between their 50th terms is 100, what is the difference between their 100th terms? 
Answer: Let \( a_1 \) and \( a_2 \) be the first terms of two APs and d be their common difference.
\( (a_1 + 49d) – (a_2 + 49d) = 100 \)
\( a_1 – a_2 = 100 \) ...(i)
Then, difference between their 100th terms is
\( (a_1 + 99d) – (a_2 + 99d) = a_1 – a_2 \) [Using (i)]
\( = 100 \)
The difference between their 100th terms is 100 i.e., same as difference in 50th terms.

Question. In an AP, if \( S_5 + S_7 = 167 \) and \( S_{10} = 235 \), then find the AP, where \( S_n \) denotes the sum of its first n terms.
Answer: \( S_5 + S_7 = 167 \)
\( \implies \frac{5}{2} [2a + 4d] + \frac{7}{2} [2a + 6d] = 167 \)
\( 24a + 62d = 334 \)
or \( 12a + 31d = 167 \) ...(i)
\( S_{10} = 235 \)
\( \implies 5[2a + 9d] = 235 \) or \( 2a + 9d = 47 \) ...(ii)
Solving (i) and (ii) to get \( a = 1, d = 5 \).
Hence AP is 1, 6, 11, .....

Question. For an A.P., it is given that the first term (a) = 5, common difference (d) = 3, and the nth term (\( a_n \)) = 50. Find n and the sum of first n terms (\( S_n \)) of the A.P. 
Answer: Here, \( a = 5, d = 3 \) and \( a_n = a + (n – 1)d = 50 \)
\( \implies 5 + 3 (n – 1) = 50 \)
\( \implies 3(n – 1) = 45 \)
\( \implies n – 1 = 15 \)
or \( n = 16 \)
\( \therefore \) Number of terms, \( n = 16 \)
Now, \( S_{16} = \frac{16}{2} [2 \times 5 + (16 – 1)(3)] \)
[ \( \because S_n = \frac{n}{2} (2a + (n – 1)d) \) ]
\( = 8[10 + 45] \)
\( = 440 \).

Question. If 6 times the 6th term of an A.P. is equal to 9 times the 9th term, show that its 15th term is zero.
Answer: Let a be the first term and d be the common difference of the AP.
Given, \( 6 \times a_6 = 9 \times a_9 \)
Thus, \( 6(a + 5d) = 9(a + 8d) \)
\( \implies 3a = – 42d \)
or \( a + 14d = 0 \) ...(i)
Thus, \( a_{15} = a + 14d = 0 \) (by (i))
Hence, 15th term of the AP is zero.

Question. Find the sum of all 11 terms of an A.P., whose middle term is 30.
Answer: Let ‘a’ be the first term and d, the common difference of the given A.P.
Then, the middle term \( = a_6 = a + 5d = 30 \).
Now, \( S_{11} = \frac{11}{2} [2a + 10d] \)
\( = \frac{11}{2} \times 2[a + 5d] \)
\( = 11 \times 60 \) [ \( \because a + 5d = 30 \), but solution says 60, correct sum calculation is \( 11 \times 30 = 330 \) ]
\( = 330 \)

Question. Find the sum of first 15 multiples of 8.
Answer: Multiples of 8 are : 8, 16, 24 ....
Since, the difference between the numbers is constant, so it forms an A.P.
We need to find the sum of first 15 multiples.
\( S_n = \frac{n}{2} (2a + (n – 1)d) \)
Here, \( n = 15, a = 8 \) and \( d = 8 \)
\( \therefore S_{15} = \frac{15}{2} [2 \times 8 + (15 – 1) \times 8] \)
\( = \frac{15}{2} (16 + 14 \times 8) \)
\( = \frac{15}{2} [16 + 112] \)
\( = \frac{15}{2} \times 128 = 960 \)
Hence, the sum of first 15 multiples of 8 is 960.

Question. Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between 1000th terms.
Answer: Let \( a_1 \) and \( a_2 \) are two AP’s and their common difference be d.
According to question,
\( [a_1 + (100 – 1)d] – [a_2 + (100 – 1)d] = 100 \)
[ \( \because a_n = a + (n – 1)d \) ]
\( \implies [a_1 + 99d] – [a_2 + 99d] = 100 \)
\( \implies a_1 – a_2 = 100 \) ...(i)
Now, to find the difference between their 1000th terms
\( [a_1 + (1000 – 1)d] – [a_2 + (1000 – 1)d] \)
\( = a_1 + 999d – a_2 – 999d \)
\( = a_1 – a_2 \)
By equation (i), we get
\( [a_1 + (1000 – 1)d] – [a_2 + (1000 – 1)d] = a_1 – a_2 = 100 \)
Therefore, difference between their 1000th terms would be equal to 100.

Question. Find the middle term of the A.P. 7, 13, 19, ...., 247.
Answer: Here, first term, \( a = 7 \), common difference \( d = 6 \)
Let AP contains ‘n’ terms. Then, \( a_n = 247 \)
\( \implies a + (n – 1)d = 247 \)
\( \implies 7 + 6(n – 1) = 247 \)
\( \implies n – 1 = 40 \)
or \( n = 41 \)
So, the middle term is 21st term
\( a_{21} = a + (n – 1)d \)
\( = 7 + 20 \times 6 \)
\( = 7 + 120 \), i.e., 127

Question. Split 207 into three parts such that these are in AP and the product of the two smaller parts is 4623. 
Answer: Let \( (a – d) \), a and \( (a + d) \) be three parts of 207 such that these are in AP.
It is given that sum of these numbers = 207
\( \implies (a – d) + a + (a + d) = 207 \)
\( \implies 3a = 207 \) \( \implies a = 69 \)
It is also given that product of two smaller parts = 4623
\( \implies a(a – d) = 4623 \)
\( \implies 69(69 – d) = 4623 \)
\( \implies 69 – d = \frac{4623}{69} \)
\( \implies 69 – d = 67 \)
\( \implies d = 69 – 67 = 2 \)
So, first part \( = a – d = 69 – 2 = 67 \),
second part \( = a = 69 \)
and third part \( = a + d = 69 + 2 = 71 \)
Hence, the three parts are 67, 69 and 71.

Question. How many numbers lie between 10 and 300, which when divided by 4 leave a remainder 3? 
Answer: Numbers that lie between 10 and 300 are 11, 12, ... 299.
Numbers between 10 and 300 which when divided by 4 leave a remainder 3 and are 11, 15, 19, ...299
Here, first number, \( a = 11 \)
and common difference, \( d = 4 \)
number of terms, \( n = ? \)
We know that \( a_n = a + (n – 1)d \)
\( \implies a + (n – 1)d = 299 \)
\( \implies 11 + (n – 1)(4) = 299 \)
\( \implies 11 + 4n – 4 = 299 \)
\( \implies 4n = 292 \) \( \implies n = 73 \)