CBSE Class 10 Mathematics Arithmetic Progressions VBQs Set C

Arithmetic Progression

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, \( n^{th} \) term, \( a_n = a + (n – 1)d \), where, ‘n’ is the number of the terms
\( \Rightarrow 99 = 12 + (n – 1) 3 \)
\( \Rightarrow 99 = 12 + 3n – 3 \)
\( \Rightarrow 3n = 90 \)
\( \Rightarrow n = 30 \)
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, \( a_7 = a + (n – 1)d \)
\( \Rightarrow 4 = a + (7 – 1) (– 4) \)
\( \Rightarrow 4 = a – 24 \)
\( \Rightarrow a = 28 \)
Hence, the first term of the A.P. is 28.

Question. Write the \( n^{th} \) 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, \( d = \frac{1+m}{m} - \frac{1}{m} = \frac{1+m-1}{m} = \frac{m}{m} = 1 \)
\( \therefore n^{th} \) 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 \( n^{th} \) 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, ......
\( n^{th} \) term, \( a_n = 129 \)
Here, first term, \( a = –1 \)
Common difference, \( d = 4 – (–1) = 5 \)
\( \therefore a + (n – 1)d = 129 \)
\( \Rightarrow – 1 + (n – 1)5 = 129 \)
\( \Rightarrow – 1 + 5n – 5 = 129 \)
\( \Rightarrow 5n – 6 = 129 \)
\( \Rightarrow 5n = 135 \)
\( \Rightarrow n = 27 \)
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 \( a_n = a + (n – 1)d \)
\( \Rightarrow a_{30} = a + (30 – 1)d = a + 29d \)
\( \Rightarrow a_{20} = a + (20 – 1)d = a + 19d \)
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)]
\( a_{30} – a_{20} = –40 \)

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.
\( \Rightarrow c = \frac{b + 2b}{2} = \frac{3b}{2} \)
\( \therefore b : c = 2 : 3 \)

Question. What is the common difference of an A.P. in which \( a_{21} – a_{7} = 84 \)? 
Answer: Let a be 1st term and d be the common difference.
\( a_{21} - a_7 = 84 \)
\( a + (21-1)d - [a + (7-1)d] = 84 \)
\( a + 20d - a - 6d = 84 \)
\( 14d = 84 \)
\( d = 6 \)
common difference is 6.

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

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, \( n^{th} \) 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 = \( \frac{1 + 2 + 3 + ... + 11}{11} = \frac{\frac{11(11 + 1)}{2 \times 11}}{11} \)
\( = \frac{11 \times 12}{2 \times 11} = 6 \)
\( \therefore \) 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
\( 2(a^2 + b^2) = (a – b)^2 + (a + b)^2 \)
R.H.S. = \( (a^2 + b^2 - 2ab) + (a^2 + b^2 + 2ab) \)
\( = 2(a^2 + b^2) = \) L.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
\( a + 16d = a + 9d + 7 \)
\( \Rightarrow 7d = 7 \Rightarrow d = 1 \)
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 \( a_n = a + (n – 1)d \)
\( \Rightarrow 204 = 12 + (n – 1) \times 4 \)
\( \Rightarrow 204 = 12 + 4n – 4 \)
\( \Rightarrow 4n = 204 – 8 \)
\( \Rightarrow n = \frac{196}{4} = 49 \)
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., \( a + (3 – 1)d = 16 \)
\( \Rightarrow a + 2d = 16 \) ...(i)
and \( a_7 = a_5 + 12 \) (given)
\( a + (7 – 1)d = a + (5 – 1)d + 12 \)
\( \Rightarrow a + 6d = a + 4d + 12 \)
\( \Rightarrow 2d = 12 \)
\( \Rightarrow d = 6 \)
If we put the value of d in equation (i), we get
\( a + 2 \times 6 = 16 \)
\( \Rightarrow a = 4 \)
\( \therefore \) 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 APs
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
\( b_n – a_n = [7 + (n – 1)d] – [2 + (n – 1)d] \)
\( = 7 + (n – 1)d – 2 – (n – 1)d = 5 \)
\( \Rightarrow b_n – a_n = 5 \)
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
\( a_n = 243 + 120 = 363 \)
\( \Rightarrow a + (n – 1)d = 363 \)
\( \Rightarrow 3 + (n – 1) \times 12 = 363 \)
\( \Rightarrow (n – 1) = \frac{360}{12} = 30 \)
\( \Rightarrow n = 31 \)
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 \( n^{th} \) 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 \( n^{th} \) 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 \( n^{th} \) 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.
\( S_n = 3 + 6 + 9 + 12 + ...... + 24 \)
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, \( 7(a + 6d) = 11(a + 10d) \) [\( \because a_n = a + (n – 1)d \)]
\( \Rightarrow 7a + 42d = 11a + 110d \)
\( \Rightarrow 7a – 11a = 110d – 42d \)
\( \Rightarrow – 4a = 68d \)
\( \Rightarrow a = – 17d \)
Now, 18th term of A.P.
\( a_{18} = a + (18 – 1)d = a + 17d \)
putting \( a = – 17d \)
\( = –17d + 17d = 0 \)
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
\( a + 9d = – 4 \)
\( a + 21d = – 16 \)
\( – – + \)
\( – 12d = 12 \Rightarrow d = – 1 \)
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) = \( a + (n – 1)d \)
\( 496 = 208 + (n – 1) \times 8 \)
\( \Rightarrow (n – 1) \times 8 = 496 – 208 = 288 \)
\( \Rightarrow (n – 1) = \frac{288}{8} = 36 \)
\( \Rightarrow n = 37 \)
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, \( a_3 = a + (3 – 1)d \Rightarrow 5 = a + 2d \) ...(i)
and \( a_7 = a + (7 – 1)d \Rightarrow 9 = a + 6d \) ...(ii)
On solving (i) and (ii), we get
\( 4d = 4 \Rightarrow d = 1 \)
If we put the value of ‘d’ in equation (i), we get:
\( 5 = a + 2 \times 1 \Rightarrow a = 3 \)
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, \( S_9 = S_{11} \)
\( \Rightarrow \frac{9}{2} [2a + (9 – 1)d] = \frac{11}{2} [2a + (11 – 1)d] \)
\( [ \because S_n = \frac{n}{2} [2a + (n – 1)d] ] \)
\( \Rightarrow 9[2a + 8d] = 11[2a + 10d] \)
\( \Rightarrow 18a + 72d = 22a + 110d \)
\( \Rightarrow 4a = – 38d \Rightarrow 2a = – 19d \) ...(i)
Now, sum of first 20 terms:
\( S_{20} = \frac{20}{2} [2a + (20 – 1)d] \)
\( = 10[– 19d + 19d] \) [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: Numbers divisible by both 2 and 5 must be divisible by 10. The sequence is 110, 120, 130, … , 990.
\( a_n = 990 \)
\( \Rightarrow 110 + (n - 1) \times 10 = 990 \)
\( \therefore n = 89 \)

Question. For what value of n, are the \( n^{th} \) terms of two A.Ps 63, 65, 67,... and 3, 10, 17,... equal? 
Answer: Let a, d and A, D be the 1st term and common difference of the 2 A.Ps respectively. n is same.
\( a = 63, d = 2 \)
\( A = 3, D = 7 \)
\( a_n = A_n \)
\( \Rightarrow a + (n-1)d = A + (n-1)D \)
\( \Rightarrow 63 + (n-1)2 = 3 + (n-1)7 \)
\( \Rightarrow 63 + 2n - 2 = 3 + 7n - 7 \)
\( \Rightarrow 61 + 2n = 7n - 4 \)
\( \Rightarrow 65 = 5n \Rightarrow 13 = n \)
Hence, when n is 13, the \( n^{th} \) terms are equal i.e., \( a_{13} = A_{13} \).

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: \( S_5 + S_7 = 167 \)
\( \Rightarrow \frac{5}{2} [2a + 4d] + \frac{7}{2} [2a + 6d] = 167 \)
\( 24a + 62d = 334 \) or \( 12a + 31d = 167 \) ...(i)
\( S_{10} = 235 \)
\( \Rightarrow 5[2a + 9d] = 235 \) or \( 2a + 9d = 47 \) ...(ii)
Solving (i) and (ii) to get \( a = 1, d = 5 \).
Hence AP is 1, 6, 11, .....

Question. The 4th term of an A.P. is zero. Prove that the 25th term of the A.P. is three times its 11th term. 
Answer: We have,
\( a_4 = 0 \)
\( a + 3d = 0 \) [\( a + (n-1)d = a_n \)]
\( 3d = -a \)
or \( -3d = a \) ...(1)
Now, \( a_{25} = a + 24d \) [\( a + (n-1)d = a_n \)]
\( = -3d + 24d \) (Putting value of 'a' from eq (1))
\( = 21d \) ...(2)
\( a_{11} = a + 10d \)
\( = -3d + 10d \)
\( = 7d \) ...(3) ( \( a = -3d \) )
From eq (2) & eq (3)
\( a_{25} = 3(a_{11}) \)
Hence Proved.

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

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, \( 6(a + 5d) = 9(a + 8d) \)
\( \Rightarrow 6a + 30d = 9a + 72d \)
\( \Rightarrow 3a = -42d \)
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, \( S_{11} = \frac{11}{2} [2a + 10d] \)
\( = \frac{11}{2} \times 2 [a + 5d] \)
\( = 11 \times 30 \) [\( \because a + 5d = 30 \)]
\( = 330 \)

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 their 1000th terms. 
Answer: Let \( a_1 \) and \( a_2 \) are two AP’s and their common difference be d.
According to question,
\( [a_1 + (100 - 1)d] - [a_2 + (100 - 1)d] = 100 \)
[\( \because a_n = a + (n - 1)d \)]
\( \Rightarrow [a_1 + 99d] - [a_2 + 99d] = 100 \)
\( \Rightarrow a_1 - a_2 = 100 \) ...(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
\( a_1 - a_2 = 100 \)
Therefore, difference between their 1000th terms would be equal to 100.

SHORT ANSWER (SA-II) Type Questions 

Question. Justify whether it is true to say that the following are the nth terms of an AP: (i) 2n – 3 (ii) 3n² + 5 (iii) 1 + n + n² 
Answer: (i) Yes, \( (2n - 3) \) is the nth term of an A.P.
It is given that \( a_n = 2n - 3 \)
Put \( n = 1, a_1 = 2(1) - 3 = 2 - 3 = -1 \)
\( n = 2, a_2 = 2(2) - 3 = 4 - 3 = 1 \)
\( n = 3, a_3 = 2(3) - 3 = 6 - 3 = 3 \)
\( n = 4, a_4 = 2(4) - 3 = 8 - 3 = 5 \)
List of numbers becomes -1, 1, 3, 5...
Here, \( a_2 - a_1 = 1 - (-1) = 2 \)
\( a_3 - a_2 = 3 - 1 = 2 \)
\( a_4 - a_3 = 5 - 3 = 2 \)
Clearly, \( a_2 - a_1 = a_3 - a_2 = a_4 - a_3 \)
Hence, \( 2n - 3 \) is the nth term of an AP.

(ii) No, \( (3n^2 + 5) \) is not the nth term of an AP.
It is given that \( a_n = 3n^2 + 5 \)
Put \( n = 1, a_1 = 3(1)^2 + 5 = 3 + 5 = 8 \)
\( n = 2, a_2 = 3(2)^2 + 5 = 12 + 5 = 17 \)
\( n = 3, a_3 = 3(3)^2 + 5 = 27 + 5 = 32 \)
List of number becomes 8, 17, 32 ...
Here, \( a_2 - a_1 = 17 - 8 = 9 \),
\( a_3 - a_2 = 32 - 17 = 15 \)
Clearly, \( a_2 - a_1 \neq a_3 - a_2 \)
Hence, \( (3n^2 + 5) \) is not the nth term of an AP.

(iii) No, \( (1 + n + n^2) \) is not the nth term of an AP.
It is given that \( a_n = 1 + n + n^2 \)
Put \( n = 1, a_1 = 1 + (1) + (1)^2 = 1 + 1 + 1 = 3 \)
\( n = 2, a_2 = 1 + (2) + (2)^2 = 1 + 2 + 4 = 7 \)
\( n = 3, a_3 = 1 + (3) + (3)^2 = 1 + 3 + 9 = 13 \)
List of number becomes 3, 7, 13 ...
Here, \( a_2 - a_1 = 7 - 3 = 4 \)
\( a_3 - a_2 = 13 - 7 = 6 \)
Clearly, \( a_2 - a_1 \neq a_3 - a_2 \)
Hence, \( (1 + n + n^2) \) is not the nth term of an AP.

Question. Find a, b and c such that the following numbers in AP: a, 7, b, 23, c. 
Answer: It is given that a, 7, b, 23, c are in AP
They have a common difference
i.e., \( 7 - a = b - 7 = 23 - b = c - 23 \)
Taking second and third terms, we get
\( b - 7 = 23 - b \Rightarrow 2b = 30 \Rightarrow b = 15 \)
Taking first and second terms, we get
\( 7 - a = b - 7 \)
\( \Rightarrow 7 - a = 15 - 7 \) [As \( b = 15 \)]
\( \Rightarrow 7 - a = 8 \Rightarrow a = -1 \)
Taking third and fourth terms, we get
\( 23 - b = c - 23 \)
\( \Rightarrow 23 - 15 = c - 23 \) [As \( b = 15 \)]
\( \Rightarrow 8 = c - 23 \Rightarrow c = 31 \)
Hence, \( a = -1, b = 15 \) and \( c = 31 \).

Question. Determine the AP whose 5th term is 19 and the difference of the 8th term from the 13th term is 20. 
Answer: Let the first term of an AP be a and common difference be d.
It is given that \( a_5 = 19 \)
and \( a_{13} - a_8 = 20 \)
We know that \( a_n = a + (n - 1)d \)
\( a_5 = a + (5 - 1)d = a + 4d = 19 \) ...(i)
Also, \( a_{13} - a_8 = 20 \)
\( \Rightarrow (a + 12d) - (a + 7d) = 20 \)
\( \Rightarrow a + 12d - a - 7d = 20 \)
\( \Rightarrow 5d = 20 \Rightarrow d = 4 \)
Putting the value of \( d = 4 \) in equation (i), we get
\( a + 4(4) = 19 \)
\( \Rightarrow a + 16 = 19 \)
\( \Rightarrow a = 3 \)
So, the required AP will be:
a, a + d, a + 2d, a + 3d ...
i.e., 3, 3 + 4, 3 + 2(4), 3 + 3(4) ...
i.e., 3, 7, 11, 15 ...

Question. The sum of the first 30 terms of an A.P. is 1920. If the fourth term is 18, find its 11th term. 
Answer: Let a be the first term and d, the common difference of the AP
Here, number of terms, \( n = 30 \)
Then, \( S_{30} = \frac{30}{2} [2a + 29d] = 1920 \)
or \( 2a + 29d = 128 \) ...(i)
Also, \( a_4 = a + 3d = 18 \) ...(ii)
On solving equations (i) and (ii), we get:
\( d = 4 \) and \( a = 6 \)
Thus, \( a_{11} = a + 10d = 6 + 40 = 46 \).
Hence, 11th term of the AP is 46.

Question. Which term of the A.P. \( 20, 19\frac{1}{4}, 18\frac{1}{2}, 17\frac{3}{4}, \dots \) is the first negative term.
Answer: Here, first term \( a = 20 \) and common difference \( d = -\frac{3}{4} \)
Let \( a_n \) be the first negative term, i.e.,
\( a + (n - 1)d < 0 \)
\( \Rightarrow 20 - \frac{3}{4}(n - 1) < 0 \)
or \( 20 - \frac{3}{4}n + \frac{3}{4} < 0 \)
or \( \frac{3}{4}n > \frac{83}{4} \)
or \( n > \frac{83}{3} \)
So, the nth term is 28.

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, \( a_n = 247 \)
\( \Rightarrow a + (n - 1)d = 247 \)
\( \Rightarrow 7 + 6(n - 1) = 247 \)
\( \Rightarrow n - 1 = 40 \)
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
\( \Rightarrow (a - d) + a + (a + d) = 207 \)
\( \Rightarrow 3a = 207 \Rightarrow a = 69 \)
It is also given that product of two smaller parts = 4623
\( \Rightarrow a(a - d) = 4623 \)
\( \Rightarrow 69(69 - d) = 4623 \)
\( \Rightarrow 69 - d = \frac{4623}{69} \)
\( \Rightarrow 69 - d = 67 \)
\( \Rightarrow d = 69 - 67 = 2 \)
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.