CBSE Class 10 Maths HOTs Arithmetic Progressions Set F

Very Short Answer Type Questions

Question. Is \(-150\) a term of the A.P. \(11, 8, 5, 2, \dots\)?
Answer: Let the first term of an A.P. be \( a \) and common difference be \( d \).
We have \( a = 11, d = -3, a_n = -150 \)
Now \( a_n = a + (n - 1)d \)
\( -150 = 11 + (n - 1)(-3) \)
\( -150 = 11 - 3n + 3 \)
\( 3n = 164 \)
or, \( n = \frac{164}{3} = 54.66 \)
Since, \( 54.66 \) is not a whole number, \(-150\) is not a term of the given A.P.

Question. Find the sum of \( n \) terms of the series
\( (4 - \frac{1}{n}) + (4 - \frac{2}{n}) + (4 - \frac{3}{n}) + ....... \)
Answer: Let sum of \( n \) term be \( S_n \)
\( S_n = (4 - \frac{1}{n}) + (4 - \frac{2}{n}) + (4 - \frac{3}{n}) + ....... \) up to n terms
\( = (4 + 4 + 4 + ..... \text{ up to n terms}) + [-\frac{1}{n} - \frac{2}{n} - \frac{3}{n} - ...... \text{ up to n terms}] \)
\( = (4 + 4 + 4 + .... \text{ up to n terms}) - \frac{1}{n}(1 + 2 + 3 + ..... \text{ up to n terms}) \)
\( = 4n - \frac{1}{n} \times \frac{n(n + 1)}{2} \)
\( = 4n - \frac{n + 1}{2} = \frac{7n - 1}{2} \)
Hence, sum of \( n \) terms \( = \frac{7n - 1}{2} \)

Question. Find the number of multiple of 9 lying between 300 and 700.
Answer: The numbers, multiple of 9 between 300 and 700 are 306, 315, 324, .... 693.
Let the first term be \( a \), common difference be \( d \) and \( n^{th} \) term be \( a_n = 693 \)
\( a_n = 306 + (n - 1)9 \)
\( 693 = 306 + (n - 1)9 \)
\( (n - 1)9 = 693 - 306 = 387 \)
\( n - 1 = \frac{387}{9} = 43 \)
\( n = 43 + 1 = 44 \)
Hence there are 44 terms.

Question. If the sum of the first 14 terms of an A.P. is 1050 and its first term is 10 find its \( 20^{th} \) term.
Answer: Let the first term be \( a \), common difference be \( d \), \( n^{th} \) term be \( a_n \) and sum of \( n \) term be \( S_n \).
We have \( a = 10 \), and \( S_{14} = 1050 \)
\( S_n = \frac{n}{2}[2a + (n - 1)d] \)
\( S_{14} = \frac{14}{2}[2 \times 10 + (14 - 1)d] \)
\( 1050 = 7[20 + 13d] \)
\( 20 + 13d = \frac{1050}{7} = 150 \)
\( 13d = 130 \Rightarrow d = 10 \)
\( a_{20} = a + (n - 1)d \)
\( = 10 + 19 \times 10 = 200 \)
Hence \( a_{20} = 200 \)

Question. If the tenth term of an A.P. is 52 and the \( 17^{th} \) term is 20 more than the \( 13^{th} \) term, find A.P.
Answer: Let the first term be \( a \), common difference be \( d \) and \( n^{th} \) term be \( a_n \).
Now \( a_{10} = 52 \)
\( a + 9d = 52 \) ...(1)
Also \( a_{17} - a_{13} = 20 \)
\( (a + 16d) - (a + 12d) = 20 \)
\( 4d = 20 \)
\( d = 5 \)
Substituting this valued \( d \) in (1), we get
\( a = 7 \)
Hence AP is 7, 12, 17, 22, ...

Question. Find the sum of all odd number between 0 and 50.
Answer: Let the first term be \( a \), common difference be \( d \), \( n^{th} \)...

Question. How many multiples of 4 lie between 11 and 266?
Answer: First multiple of 4 is 12 and last multiple of 4 is 264. It forms a AP. Let multiples of 4 be \( n \).
Let the first term be \( a \), common difference be \( d \), \( n^{th} \) term be \( a_n \) and sum of \( n \) term be \( S_n \)
Here, \( a = 12, a_n = 264, d = 4 \)
\( a_n = a + (n - 1)d \)
\( 264 = 12 + (n - 1)4 \)
\( n = \frac{264 - 12}{4} + 1 \)
Hence, there are 64 multiples of 4 that lie between 11 and 266.

Question. Find the sum of the integers between 100 and 200 that are divisible by 6.
Answer: The series as per question is 102, 108, 114, ......... 198. which is an A.P.
Let the first term be \( a \), common difference be \( d \), \( n^{th} \) term be \( a_n \) and sum of \( n \) term be \( S_n \)
Here \( a = 102, d = 6 \), and \( l = 198 \)
Now \( 198 = 102 + (n - 1)6 \)
\( 96 = (n - 1)6 \)
\( \frac{96}{6} = n - 1 \)
\( n = 17 \)
Now \( S_{17} = \frac{n}{2}(a + l) \)
\( = \frac{17}{2}[102 + 198] \)

Question. Which term of the A.P. 8, 14, 20, 26,....... will be 72 more than its \( 41^{st} \) term.
Answer: Let the first term be \(a\), common difference be \(d\) and \(n^{th}\) term be \(a_n\).
We have \(a = 8, d = 6\).
Since \(n^{th}\) term is 72 more than \(41^{st}\) term. we get
\(a_n = a_{41} + 72\)
\(8 + (n - 1)6 = 8 + 40 \times 6 + 72\)
\(6n - 6 = 240 + 72\)
\(6n = 312 + 6 = 318\)
\(n = 53\)

Question. If the \(n^{th}\) term of an A.P. \(-1, 4, 9, 14, \dots\) is 129. Find the value of \(n\).
Answer: Let the first term be \(a\), common difference be \(d\) and \(n^{th}\) term be \(a_n\).
We have \(a = -1\) and \(d = 4 - (-1) = 5\)
\(-1 + (n - 1) \times 5 = a_n\)
\(-1 + 5n - 5 = 129\)
\(5n = 135\)
\(n = 27\)
Hence \(27^{th}\) term is 129.

Short Answer Type Questions

Question. In a certain A.P. \( 32^{th} \) term is twice the \( 12^{th} \) term. Prove that \( 70^{th} \) term is twice the \( 31^{st} \) term.
Answer: Let the first term of an A.P. be \( a \), common difference be \( d \) and \( n^{th} \) term be \( a_n \).
Now we have \( a_{32} = 2a_{12} \)
\( a + 31d = 2(a + 11d) \)
\( a + 31d = 2a + 22d \)
\( a = 9d \)
\( a_{70} = a + 69d \)
\( = 9d + 69d = 78d \)
\( a_{31} = a + 30d \)
\( = 9d + 30d = 39d \)
\( a_{70} = 2a_{31} \) Hence Proved.

Question. Find the middle term of the A.P. 6, 13, 20, .... 216.
Answer: Let the first term of an A.P. be \( a \), common difference be \( d \) and number of terms be \( m \).
Here, \( a = 6, a_m = 216, d = 13 - 6 = 7 \)
\( a_m = a + (m - 1)d \)
\( 216 = 6 + (m - 1)(7) \)
\( 216 - 6 = 7(m - 1) \)
\( m - 1 = \frac{210}{7} = 30 \)
\( m = 30 + 1 = 31 \)
The middle term will be \( = \frac{31 + 1}{2} = 16^{th} \)
\( a_{16} = a + (16 - 1)d \)
\( = 6 + (16 - 1)(7) \)
\( = 6 + 15 \times 7 \)
\( = 6 + 105 = 111 \)
Middle term will be 111.

Question. If the 2nd term of an A.P. is 8 and the 5th term is 17, find its 19th term.
Answer: Let the first term be \( a \) and common difference be \( d \).
Now \( a_2 = a + d \)
\( 8 = a + d \) --- (1)
and \( a_5 = a + 4d \)
\( 17 = a + 4d \) --- (2)
Solving (1) and (2), we have
\( a = 5, d = 3, \)
\( a_{19} = a + 18d \)
\( = 5 + 54 = 59 \)

Question. Find the values of \( a, b, \) and \( c \), such that the numbers \( a, 10, b, c, 31 \) are in A.P.
Answer: Let the first term be \( a \) and common difference be \( d \).
Since \( a, 10, b, c, 31 \) are in A.P.
Now \( a + d = 10 \) --- (1)
\( a + 4d = a_5 \)
\( a + 4d = 31 \) --- (2)
Solving (1) and (2) we have
\( d = 7 \) and \( a = 3 \)
Now \( a = 3, b = 3 + 14 = 17, c = 3 + 21 = 24 \)
Thus \( a = 3, b = 17, c = 24 \).

Question. For A.P. show that \( a_p + a_{p+2q} = 2a_{p+q} \).
Answer: Let the first term be \( a \) and the common difference be \( d \). Let \( a_n \) be the \( n^{th} \) term.
\( a_p = a + (p - 1)d \)
\( a_{p+2q} = a + (p + 2q - 1)d \)
\( a_p + a_{p+2q} = a + (p - 1)d + a + (p + 2q - 1)d \)
\( = a + pd - d + a + pd + 2qd - d \)
\( = 2a + 2pd + 2qd - 2d \)
or \( a_p + a_{p+2q} = 2[a + (p + q - 1)d] \) ...(1)
But \( 2a_{p+q} = 2[a + (p + q - 1)d] \) ...(2)
From (1) and (2), we get \( a_p + a_{p+2q} = 2a_{p+q} \)

Question. The sum of first \( n \) terms of an A.P. is given by \( S_n = 2n^2 + 8n \). Find the sixteenth term of the A.P.
Answer: Let the first term be \( a \), common difference be \( d \) and \( n^{th} \) term be \( a_n \).
Now \( S_n = 2n^2 + 3n \)
\( S_1 = 2 \times 1^2 + 3 \times 1 = 2 + 3 = 5 \)
Since \( S_1 = a_1 \),
\( a_1 = 5 \)
\( S_2 = 2 \times 2^2 + 3 \times 2 = 8 + 6 = 14 \)
\( a_1 + a_2 = 14 \)
\( a_2 = 14 - a_1 = 14 - 5 = 9 \)
\( d = a_2 - a_1 = 9 - 5 = 4 \)
\( a_{16} = a + (16 - 1)d \)
\( = 5 + 15 \times 4 = 65 \)

Question. Find the 20th term of an A.P. whose 3rd term is 7 and the seventh term exceeds three times the 3rd term by 2. Also find its \( n^{th} \) term (\( a_n \)).
Answer: Let the first term be \( a \), common difference be \( d \) and \( n^{th} \) term be \( a_n \).
We have \( a_3 = a + 2d = 7 \) (1)
\( a_7 = 3a_3 + 2 \)
\( a + 6d = 3 \times 7 + 2 = 23 \) (2)
Solving (1) and (2) we have
\( 4d = 16 \Rightarrow d = 4 \)
\( a + 8 = 7 \Rightarrow a = -1 \)
\( a_{20} = a + 19d = -1 + 19 \times 4 = 75 \)
\( a_n = a + (n - 1)d \)
\( = -1 + (n - 1)4 \)
\( = 4n - 5 \).
Hence \( n^{th} \) term is \( 4n - 5 \)

Question. If 7th term of an A.P. is \( \frac{1}{9} \) and 9th term is \( \frac{1}{7} \), find 63rd term.
Answer: Let the first term be \( a \), common difference be \( d \) and \( n^{th} \) term be \( a_n \).
We have \( a_7 = \frac{1}{9} \Rightarrow a + 6d = \frac{1}{9} \) (1)
\( a_9 = \frac{1}{7} \Rightarrow a + 8d = \frac{1}{7} \) (2)
Subtracting equation (1) from (2) we get
\( 2d = \frac{1}{7} - \frac{1}{9} = \frac{2}{63} \)
\( d = \frac{1}{63} \)
Substituting the value of \( d \) in (2) we get
\( a + 8 \times \frac{1}{63} = \frac{1}{7} \)
\( a = \frac{1}{7} - \frac{8}{63} = \frac{9 - 8}{63} = \frac{1}{63} \)
Thus \( a_{63} = a + (63 - 1)d \)
\( = \frac{1}{63} + 62 \times \frac{1}{63} = \frac{1 + 62}{63} \)
\( = \frac{63}{63} = 1 \)
Hence, \( a_{63} = 1 \)

Question. The ninth term of an A.P. is equal to seven times the second term and twelfth term exceeds five times the third term by 2. Find the first term and the common difference.
Answer: Let the first term be \( a \), common difference be \( d \) and \( n^{th} \) term be \( a_n \).
Now \( a_9 = 7a_2 \)
\( a + 8d = 7(a + d) \)
\( a + 8d = 7a + 7d \)
\( -6a + d = 0 \) (1)
and \( a_{12} = 5a_3 + 2 \)
\( a + 11d = 5(a + 2d) + 2 \)
\( a + 11d = 5a + 10d + 2 \)
\( -4a + d = 2 \) ...(2)
Subtracting (2) from (1), we get
\( -2a = -2 \)
\( a = 1 \)
Substituting this value of \( a \) in (1) we get
\( -6 + d = 0 \)
\( d = 6 \)
Hence first term is 1 and common difference is 6.

Question. Determine an A.P. whose third term is 9 and when fifth term is subtracted from 8th term, we get 6.
Answer: Let the first term be \( a \), common difference be \( d \) and \( n^{th} \) term be \( a_n \).
We have \( a_3 = 9 \)
\( a + 2d = 9 \) ...(1)
and \( a_8 - a_5 = 6 \)
\( (a + 7d) - (a + 4d) = 6 \)
\( 3d = 6 \)
\( d = 2 \)
Substituting this value of \( d \) in (1), we get
\( a + 2(2) = 9 \)
\( a = 5 \)
So, A.P. is 5, 7, 9, 11, ...

Question. Divide 56 in four parts in A.P. such that the ratio of the product of their extremes (1st and 4rd) to the product of means (2nd and 3rd) is 5:6.
Answer: Let the four numbers be \( a-3d, a-d, a+d, a+3d \)
Now \( a-3d + a-d + a+d + a+3d = 56 \)
\( 4a = 56 \Rightarrow a = 14 \)
Hence numbers are \( 14-3d, 14-d, 14+d, 14+3d \)
Now, according to question,
\( \frac{(14 - 3d)(14 + 3d)}{(14 - d)(14 + d)} = \frac{5}{6} \)
\( \frac{196 - 9d^2}{196 - d^2} = \frac{5}{6} \)
\( 6(196 - 9d^2) = 5(196 - d^2) \)
\( 6 \times 196 - 54d^2 = 5 \times 196 - 5d^2 \)
\( (6 - 5) \times 196 = 49d^2 \)
\( d^2 = \frac{196}{49} = 4 \)
\( d = \pm 2 \)
Thus numbers are \( a - 3d = 14 - 3 \times 2 = 8 \)
\( a - d = 14 - 2 = 12 \)
\( a + d = 14 + 2 = 16 \)
\( a + 3d = 14 + 3 \times 2 = 20 \)
Thus required AP is 8, 12, 16, 20.

Question. The \( p^{th}, q^{th} \) and \( r^{th} \) terms of an A.P. are \( a, b, \) and \( c \) respectively, Show that \( a(q - r) + b(r - p) + c(p - q) = 0 \).
Answer: Let the first term be \( A \) and the common difference be \( D \).
\( a = A + (p - 1)D \)
\( b = A + (q - 1)D \)
\( c = A + (r - 1)D \)
Now \( a(q - r) = [A + (p - 1)D](q - r) \)
\( b(r - p) = [A + (q - 1)D](r - p) \)
and \( c(p - q) = [A + (r - 1)D](p - q) \)
\( a(q - r) + b(r - p) + c(p - q) \)
\( = [A + (p - 1)D](q - r) + [A + (q - 1)D](r - p) + [A + (r - 1)D](p - q) \)
\( = A[q - r + r - p + p - q] + D[(p - 1)(q - r) + (q - 1)(r - p) + (r - 1)(p - q)] \)
\( = A[0] + D[p(q - r) - (q - r) + q(r - p) - (r - p) + r(p - q) - (p - q)] \)
\( = D[p(q - r) + q(r - p) + r(p - q)] - D[(q - r) + (r - p) + (p - q)] \)
\( = D[pq - pr + qr - qp + rp - rq] + 0 \)
\( = D[0] = 0 \)

Question. The sum of \( n \) terms of an A.P. is \( 3n^2 + 5n \). Find the A.P. Hence find its \( 15^{th} \) term.
Answer: Let the first term be \( a \), common difference be \( d \), \( n^{th} \) term be \( a_n \) and sum of \( n \) term be \( S_n \)
Now \( S_n = 3n^2 + 5n \)
\( S_{n-1} = 3(n - 1)^2 + 5(n - 1) \)
\( = 3(n^2 + 1 - 2n) + 5n - 5 \)
\( = 3n^2 + 3 - 6n + 5n - 5 \)
\( = 3n^2 - n - 2 \)
\( a_n = S_n - S_{n-1} \)
\( = 3n^2 + 5n - (3n^2 - n - 2) \)
\( = 6n + 2 \)
Thus A.P. is 8, 14, 20, .......
Now \( a_{15} = a + 14d = 8 + 14(6) = 92 \)

Question. The digit 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 then the original number. Find the number.
Answer: Let these digit of 3 digit no be \( a - d, a, a + d \)
Since their sum is 15,
\( a - d + a + a + d = 15 \)
\( 3a = 15 \Rightarrow a = 5 \)
Required 3 digit no \( = 100(a - d) + 10a + (a + d) \)
\( = 100a - 100d + 10a + a + d \)
\( = 111a - 99d \)
No obtained by reversing digit
\( = 100(a + d) + 10a + (a - d) \)
\( = 100a + 100d + 10a + a - d \)
\( = 111a + 99d \)
According the question,
\( 111a + 99d = 111a - 99d - 594 \)
\( 2 \times 99d = -594 \Rightarrow d = -3 \)
Thus number is \( 111a - 99d = 111 \times 5 - 99 \times 3 \)
\( = 555 + 297 = 852 \)

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 \( 1^{st} \) term and common difference of the 2 APs respectively.
\( n \) is same
For 1st AP, \( a = 63, d = 2 \)
For 2nd AP, \( A = 3, D = 7 \)
Since \( n^{th} \) term is same,
\( a_n = A_n \)
\( a + (n - 1)d = A + (n - 1)D \)
\( 63 + (n - 1)2 = 3 + (n - 1)7 \)
\( 63 + 2n - 2 = 3 + 7n - 7 \)
\( 61 + 2n = 7n - 4 \)
\( 65 = 5n \Rightarrow n = 13 \)
When \( n \) is 13, the \( n^{th} \) terms are equal i.e., \( a_{13} = A_{13} \)

Question. The sum of three numbers in A.P. is 12 and sum of their cubes is 288. Find the numbers.
Answer: Let the three numbers in A.P. be \( a - d, a, a + d \).
\( a - d + a + a + d = 12 \)
\( 3a = 12 \Rightarrow a = 4 \)
Also, \( (4 - d)^3 + 4^3 + (4 + d)^3 = 288 \)
\( 64 - 48d + 12d^2 - d^3 + 64 + 64 + 48d + 12d^2 + d^3 = 288 \)
\( 24d^2 + 192 = 288 \)
\( d^2 = 4 \Rightarrow d = \pm 2 \)
The numbers are 2, 4, 6 or 6, 4, 2

Question. Find the value of \( a, b, \) and \( c \) such that the numbers \( a, 7, b, 23 \) and \( c \) are in A.P.
Answer: Let the common difference be \( d \).
Since \( a, 7, b, 23 \) and \( c \) are in AP, we have
\( a + d = 7 \) ..(1)
\( a + 3d = 23 \) ..(2)
From (1) and (2), we get
\( a = -1, d = 8 \)
\( b = a + 2d = -1 + 2 \times 8 = -1 + 16 = 15 \)
\( c = a + 4d = -1 + 4 \times 8 = -1 + 32 = 31 \)
Thus \( a = -1, b = 15, c = 31 \)

Question. Find the sum of first ten multiple of 5.
Answer: Let the first term be \( a \), common difference be \( d \), \( n^{th} \) term be \( a_n \) and sum of \( n \) term be \( S_n \)
Here, \( a = 5, n = 10, d = 5 \)
\( S_n = \frac{n}{2}[2a + (n - 1)d] \)
\( S_{10} = \frac{10}{2}[2 \times 5 + (10 - 1)5] \)
\( = 5[10 + 9 \times 5] \)
\( = 5[10 + 45] \)
\( = 5 \times 55 = 275 \)
Hence the sum of first ten multiple of 5 is 275.

Question. If the sum of \( n \) terms of an A.P. is \( 2n^2 + 5n \), then find the \( 4^{th} \) term.
Answer: Let the first term be \( a \), common difference be \( d \), \( n^{th} \) term be \( a_n \) and sum of \( n \) term be \( S_n \)
Now, \( S_n = 2n^2 + 5n \)
\( n^{th} \) term of A.P.
\( a_n = S_n - S_{n-1} \)
\( a_n = (2n^2 + 5n) - [2(n - 1)^2 + 5(n - 1)] \)
\( = 2n^2 + 5n - [2(n^2 - 4n + 2) + 5n - 5] \)
\( = 2n^2 + 5n - [2n^2 + n - 3] \)
\( = 4n + 3 \)
Thus \( 4^{th} \) term \( a_4 = 4 \times 4 + 3 = 19 \)

HOTS QUESTIONS

Question. Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5.
Answer: The sequence goes like 110, 120, 130, ......... 990
Since they have a common difference of 10, they form an A.P.
Let the first term be \( a \), common difference be \( d \), \( n^{th} \) term be \( a_n \) and sum of \( n \) term be \( S_n \)
Here \( a = 110, a_n = 990, d = 10 \)
\( a_n = a + (n - 1)d \)
\( 990 = 110 + (n - 1) \times 10 \)
\( 990 - 110 = 10(n - 1) \)
\( 880 = 10(n - 1) \)
\( 88 = n - 1 \)
\( n = 88 + 1 = 89 \)
Hence, there are 89 terms between 101 and 999 divisible by both 2 and 5.

Question. How many three digit natural numbers are divisible by 7?
Answer: Let A.P. is 105, 112, 119, ........., 994 which is divisible by 7.
Let the first term be \( a \), common difference be \( d \), \( n^{th} \) term be \( a_n \) and sum of \( n \) term be \( S_n \)
Here, \( a = 105, d = 112 - 105 = 7, a_n = 994 \), then
\( a_n = a + (n - 1)d \)
\( 994 = 105 + (n - 1) \times 7 \)
\( 889 = (n - 1) \times 7 \)
\( n - 1 = \frac{889}{7} = 127 \)
\( n = 127 + 1 = 128 \)
Hence, there 128 terms divisible by 7 in A.P.

Question. How many two digit numbers are divisible by 7?
Answer: Two digit numbers which are divisible by 7 are 14, 21, 28, ..... 98. It forms an A.P.
Let the first term be \( a \), common difference be \( d \), \( n^{th} \) term be \( a_n \) and sum of \( n \) term be \( S_n \)
Here \( a = 14, d = 7, a_n = 98 \)
Now \( a_n = a + (n - 1)d \)
\( 98 = 14 + (n - 1)7 \)
\( 98 - 14 = 7n - 7 \)
\( 84 + 7 = 7n \)
\( 7n = 91 \implies n = 13 \)

Question. How many three digit numbers are such that when divided by 7, leave a remainder 3 in each case?
Answer: When a three digit number divided by 7 and leave 3 as remainder are 101, 108, 115, ..... 997
These are in A.P.
Let the first term be \( a \), common difference be \( d \), \( n^{th} \) term be \( a_n \) and sum of \( n \) term be \( S_n \)
Here \( a = 101, d = 7, a_n = 997 \)
Now \( a_n = a + (n - 1)d \)
\( 997 = 101 + (n - 1)7 \)
\( 997 - 101 = 896 = (n - 1)7 \)
\( \frac{896}{7} = n - 1 \)
\( n = 128 + 1 = 129 \)
Hence, 129 numbers are divided by 7 which leaves remainder is 3.

Question. Prove that the \( n^{th} \) term of an A.P. can not be \( n^2 + 1 \). Justify your answer.
Answer: Let \( n^{th} \) term of A.P.
\( a_n = n^2 + 1 \)
Substituting the value of \( n = 1, 2, 3, ..... \) we get
\( a_1 = 1^2 + 1 = 2 \)
\( a_2 = 2^2 + 1 = 5 \)
\( a_3 = 3^2 + 1 = 10 \)
The obtained sequence is 2, 5, 10, 17,......
Its common difference
\( a_2 - a_1 = a_3 - a_2 = a_4 - a_3 \)
\( 5 - 2 \neq 10 - 5 \neq 17 - 10 \)
\( 3 \neq 5 \neq 7 \)
Since the sequence has no. common difference, \( n^2 + 1 \) is not a form of \( n^{th} \) term of an A.P.

Question. Find the sum of all two digits odd positive numbers.
Answer: The list of 2 digits odd positive numbers are 11, 13 ...... 99. It forms an AP.
Let the first term be \( a \), common difference be \( d \), \( n^{th} \) term be \( a_n \) and sum of \( n \) term be \( S_n \)
Here \( a = 11, d = 2, l = 99 \)
Now \( a_n = a + (n - 1)d \)
\( 99 = 11 + (n - 1)2 \)
\( 88 = (n - 1)2 \)
\( n = 44 + 1 = 45 \)
\( S_n = \frac{n}{2}[a + a_n] \)
\( = \frac{45}{2}[11 + 99] \)
\( S_n = \frac{45 \times 110}{2} = 2475 \)
Hence the sum of given A.P. is \( S_n = 2475 \)