CBSE Class 10 Arithmetic Progressions Sure Shot Questions Set I

Question. Find the sum of the odd numbers between 0 and 50.
Answer: Odd numbers between 0 and 50 are:
1, 3, 5, 7, ....., 49
These numbers are in A.P. such that
\( a = 1 \) and \( l = 49 \)
Here, \( d = 3 - 1 = 2 \)
\( \therefore T_n = a + (n - 1) d \)
\( \Rightarrow 49 = 1 + (n - 1) 2 \)
\( \Rightarrow 49 - 1 = (n - 1) 2 \)
\( \Rightarrow (n - 1) = \frac{48}{2} = 24 \)
\( \therefore n = 24 + 1 = 25 \)
Now, \( S_{25} = \frac{25}{2} [1 + 49] \)
\( = \frac{25}{2} [50] \)
\( = 25 \times 25 = 625 \)
Thus, the sum of odd numbers between 0 and 50 is 625.

Question. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹ 200 for the first day, ₹ 250 for the second day, ₹ 300 for the third day, etc., the penalty for each succeeding day being ₹ 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?
Answer: Here, penalty for delay on
1st day = ₹ 200
2nd day = ₹ 250
3rd day = ₹ 300
...............
...............
Now, 200, 250, 300, ..... are in A.P. such that
\( a = 200, d = 250 - 200 = 50 \)
\( \therefore S_{30} \) is given by
\( S_{30} = \frac{30}{2} [2 (200) + (30 - 1) \times 50] \) [using \( S_n = \frac{n}{2} [2a + (n - 1)d] \)]
\( = 15 [400 + 29 \times 50] \)
\( = 15 [400 + 1450] \)
\( = 15 \times 1850 = 27,750 \)
Thus, penalty for the delay for 30 days is ₹ 27,750.

Question. A sum of ₹ 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹ 20 less than its preceding prize, find the value of each of the prizes.
Answer: Sum of all the prizes = ₹ 700
Let the first prize = \( a \)
\( \therefore \) 2nd prize = \( (a - 20) \)
3rd prize = \( (a - 40) \)
4th prize = \( (a - 60) \)
........................................
Thus, we have, first term = \( a \)
Common difference \( d = -20 \)
Number of prizes, \( n = 7 \)
Sum of 7 terms \( S_n = 700 \)
Since, \( S_n = \frac{n}{2} [2a + (n - 1) d] \)
\( \Rightarrow 700 = \frac{7}{2} [2 (a) + (7 - 1) \times (- 20)] \)
\( \Rightarrow 700 = \frac{7}{2} [2a + (6 \times - 20)] \)
\( \Rightarrow 700 \times \frac{2}{7} = 2a - 120 \)
\( \Rightarrow 200 = 2a - 120 \)
\( \Rightarrow 2a = 200 + 120 = 320 \)
\( \Rightarrow a = \frac{320}{2} = 160 \)
Thus, the values of the seven prizes are:
₹ 160, ₹ (160 - 20), ₹ (160 - 40), ₹ (160 - 60), ₹ (160 - 80), ₹ (160 - 100) and ₹ (160 - 120)
\( \Rightarrow \) ₹ 160, ₹ 140, ₹ 120, ₹ 100, ₹ 80, ₹ 60 and ₹ 40.

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 studying, 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 classes = 12
Each class has 3 sections.
\( \therefore \) Number of plants planted by class I = \( 1 \times 3 = 3 \)
Number of plants planted by class II = \( 2 \times 3 = 6 \)
Number of plants planted by class III = \( 3 \times 3 = 9 \)
Number of plants planted by class IV = \( 4 \times 3 = 12 \)
.......................................................................................................
Number of plants planted by class XII = \( 12 \times 3 = 36 \)
The numbers 3, 6, 9, 12, ..........., 36 are in A.P.
Here, \( a = 3 \) and \( d = 6 - 3 = 3 \)
Number of classes \( n = 12 \)
\( \therefore \) Sum of the \( n \) terms of the above A.P., is given by
\( S_{12} = \frac{12}{2} [2 (3) + (12 - 1) 3] \) [using \( S_n = \frac{n}{2} [2a + (n - 1)d] \)]
\( = 6 [6 + 11 \times 3] \)
\( = 6 [6 + 33] \)
\( = 6 \times 39 = 234 \)
Thus, the total number of trees = 234.

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: We have:
The number of logs:
1st row = 20
2nd row = 19
3rd row = 18
obviously, the numbers 20, 19, 18, ....., are in A.P. such that
\( a = 20, d = 19 - 20 = - 1 \)
Let the numbers of rows be \( n \).
\( \therefore S_n = 200 \)
Now, using \( S_n = \frac{n}{2} [2a + (n - 1) d] \), we get
\( S_n = \frac{n}{2} [2 (20) + (n - 1) \times (- 1)] \)
\( \Rightarrow 200 = \frac{n}{2} [40 - (n - 1)] \)
\( \Rightarrow 2 \times 200 = n \times [40 - n + 1] \)
\( \Rightarrow 400 = n \times [41 - n] \)
\( \Rightarrow 400 = 41n - n^2 \)
\( \Rightarrow n^2 - 41n + 400 = 0 \)
\( \Rightarrow n^2 - 16n - 25n + 400 = 0 \)
\( \Rightarrow n (n - 16) - 25 (n - 16) = 0 \)
\( \Rightarrow (n - 16) (n - 25) = 0 \)
Either \( n - 16 = 0 \Rightarrow n = 16 \)
or \( n - 25 = 0 \Rightarrow n = 25 \)
If \( n = 25 \), then \( T_{25} = a + (25 - 1) d = 20 + 24(-1) = 20 - 24 = -4 \)
But number of logs cannot be negative.
\( \therefore n = 25 \) is not required.
Thus, \( n = 16 \)
Number of rows = 16
Now, \( T_{16} = a + (16 - 1) d \)
\( = 20 + 15 \times (- 1) \)
\( = 20 - 15 = 5 \)
Number of logs in the 16th (top) row is 5.

Question. Which term of the A.P.: 121, 117, 113, ..., is its first negative term?
Answer: We have the A.P. having \( a = 121 \) and \( d = 117 - 121 = - 4 \)
\( \therefore a_n = a + (n - 1) d \)
\( = 121 + (n - 1) \times (- 4) \)
\( = 121 - 4n + 4 \)
\( = 125 - 4n \)
For the first negative term, we have \( a_n < 0 \)
\( \Rightarrow (125 - 4n) < 0 \)
\( \Rightarrow 125 < 4n \)
\( \Rightarrow \frac{125}{4} < n \)
\( \Rightarrow 31 \frac{1}{4} < n \)
or \( n > 31 \frac{1}{4} \)
Thus, the first negative term is 32nd term.

Question. The sum of the third and the seventh terms of an A.P. is 6 and their product is 8. Find the sum of first sixteen terms of the A.P..
Answer: Here, \( T_3 + T_7 = 6 \) and \( T_3 \times T_7 = 8 \)
Let the first term = \( a \) and the common difference = \( d \)
\( \therefore T_3 = a + 2d \) and \( T_7 = a + 6d \)
\( T_3 + T_7 = 6 \)
\( \therefore (a + 2d) + (a + 6d) = 6 \)
\( \Rightarrow 2a + 8d = 6 \)
\( \Rightarrow a + 4d = 3 \) ...(1)
Again \( T_3 \times T_7 = 8 \)
\( \therefore (a + 2d) \times (a + 6d) = 8 \)
\( \Rightarrow (a + 4d - 2d) \times (a + 4d + 2d) = 8 \)
\( \Rightarrow [(a + 4d) - 2d] \times [(a + 4d) + 2d] = 8 \)
\( \Rightarrow [3 - 2d] \times [3 + 2d] = 8 \) [From (1)]
\( \Rightarrow 3^2 - (2d)^2 = 8 \)
\( \Rightarrow 9 - 4d^2 = 8 \)
\( \Rightarrow - 4d^2 = 8 - 9 = - 1 \)
\( \Rightarrow d^2 = \frac{- 1}{- 4} = \frac{1}{4} \)
\( \Rightarrow d = \pm \frac{1}{2} \)
When \( d = \frac{1}{2} \).
From (1), we have:
\( a + 4 \left( \frac{1}{2} \right) = 3 \)
\( \Rightarrow a + 2 = 3 \) or \( a = 3 - 2 = 1 \)
Now, Using \( S_n = \frac{n}{2} [2a + (n - 1) d] \), we get
\( S_{16} = \frac{16}{2} [2 (1) + (16 - 1) \times \frac{1}{2}] \)
\( = 8 [2 + \frac{15}{2}] \)
\( = 16 + 60 = 76 \)
i.e., the sum of first 16 terms = 76
When \( d = - \frac{1}{2} \).
From (1), we have:
\( a + 4 \left( - \frac{1}{2} \right) = 3 \)
\( \Rightarrow a - 2 = 3 \Rightarrow a = 5 \)
Again, the sum of first sixteen terms
\( S_{16} = \frac{16}{2} [2 (5) + (16 - 1) \times (- \frac{1}{2})] \)
\( = 8 [10 + (- \frac{15}{2})] \)
\( = 80 - 60 = 20 \)
i.e., the sum of first 16 terms = 20

Question. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.
Answer: We have, the following consecutive numbers on the houses of a row;
1, 2, 3, 4, 5, ....., 49.
These numbers are in an A.P., such that
\( a = 1, d = 2 - 1 = 1, n = 49 \)
Let one of the houses be numbered as x
\( \therefore \) Number of houses preceding it = \( x - 1 \)
Number of houses following it = \( 49 - x \)
Now, the sum of the house-numbers preceding x is given by:
Using \( S_n = \frac{n}{2} [2a + (n - 1) d] \)
\( S_{x - 1} = \frac{x - 1}{2} [2 (1) + (x - 1 - 1) \times 1] \)
\( = \frac{x - 1}{2} [2 + x - 2] \)
\( = \frac{x - 1}{2} [x] = \frac{x(x - 1)}{2} = \frac{x^2}{2} - \frac{x}{2} \)
The houses beyond x are numbered as \( (x + 1), (x + 2), (x + 3), ....., 49 \)
\( \therefore \) For these house numbers (which are in an A.P.),
First term (a) = \( x + 1 \)
Last term (l) = 49
\( \therefore \) Using \( S_n = \frac{n}{2} [a + l] \), we have
\( S_{49 - x} = \frac{49 - x}{2} [(x + 1) + 49] \)
\( = \frac{49 - x}{2} [x + 50] \)
\( = \frac{49x}{2} - \frac{x^2}{2} + \frac{49 \times 50}{2} - 25x \)
\( = \frac{49x}{2} - 25x - \frac{x^2}{2} + (49 \times 25) \)
\( = \left( \frac{49x - 50x}{2} \right) - \frac{x^2}{2} + (49 \times 25) \)
\( = -\frac{x}{2} - \frac{x^2}{2} + (49 \times 25) \)
According to the question,
[Sum of house numbers preceding x] = [Sum of house numbers following x]
i.e., \( S_{x - 1} = S_{49 - x} \)
\( \Rightarrow \frac{x^2}{2} - \frac{x}{2} = -\frac{x}{2} - \frac{x^2}{2} + (49 \times 25) \)
\( \Rightarrow \left( \frac{x^2}{2} + \frac{x^2}{2} \right) - \frac{x}{2} + \frac{x}{2} = (49 \times 25) \)
\( \Rightarrow \frac{2x^2}{2} = (49 \times 25) \)
\( \Rightarrow x^2 = (49 \times 25) \)
\( \Rightarrow x = \pm \sqrt{49 \times 25} \)
\( \Rightarrow x = \pm (7 \times 5) = \pm 35 \)
But x cannot be taken as -ve
\( \therefore x = 35 \)

MORE QUESTIONS-

I. VERY SHORT ANSWER TYPE QUESTIONS

Question. If the numbers \( x - 2, 4x - 1 \) and \( 5x + 2 \) are in A.P. Find the value of x.
Answer: \( \because x - 2, 4x - 1 \) and \( 5x + 2 \) are in A.P.
\( \therefore (4x - 1) - (x - 2) = (5x + 2) - (4x - 1) \)
\( \Rightarrow 3x + 1 = x + 3 \)
\( \Rightarrow 2x = 2 \Rightarrow x = 1 \)

Question. Which term of the A.P. 4, 9, 14, ..... is 109?
Answer: Let 109 is the nth term,
\( \therefore \) Using \( T_n = a + (n - 1) d \), we have:
\( 109 = 4 + (n - 1) 5 \) [\( \because a = 4 \) and \( d = 9 - 4 = 5 \)]
\( 105 = (n - 1) 5 \)
\( n - 1 = 21 \Rightarrow n = 22 \)

\( \Rightarrow n - 1 = \frac{109 - 4}{5} = \frac{105}{5} = 21 \)
\( \Rightarrow n = 21 + 1 = 22 \)
Thus, the 22nd term is 109.

Question. If a, (a − 2) and 3a are in A.P. then what is the value of a?
Answer: \( \because \) a, (a − 2) and 3a are in A.P.
\( \therefore (a - 2) - a = 3a - (a - 2) \)
\( \Rightarrow a - 2 - a = 3a - a + 2 \)
\( \Rightarrow - 2 = 2a + 2 \)
\( \Rightarrow 2a = - 2 - 2 = - 4 \)
\( \Rightarrow a = \frac{- 4}{2} = - 2 \)
Thus, the required value of a is − 2.

Question. How many terms are there in the A.P.? 7, 10, 13, ....., 151
Answer: Here, \( a = 7, d = 10 - 7 = 3 \)
Let there are n-terms.
\( \therefore T_n = a + (n - 1) d \)
\( \Rightarrow 151 = 7 + (n - 1) \times 3 \)
\( \Rightarrow \frac{151 - 7}{3} = n - 1 \)
\( \Rightarrow \frac{144}{3} = n - 1 \Rightarrow n = 48 + 1 = 49 \)
i.e., n = 49

Question. Which term of the A.P. 72, 63, 54, ..... is 0?
Answer: Here, \( a = 72, d = 63 - 72 = - 9 \)
Let nth term of this A.P. be 0
\( \therefore T_n = a + (n - 1) d \)
\( \Rightarrow 72 + (n - 1) \times (- 9) = 0 \)
\( \Rightarrow (n - 1) = \frac{- 72}{- 9} = 8 \)
\( \Rightarrow n = 8 + 1 = 9 \)
Thus the 9th term of the A.P. is 0.

Question. The first term of an A.P. is 6 and its common difference is − 2. Find its 18th term.
Answer: Using \( T_n = a + (n - 1) d \), we have:
\( T_{18} = 6 + (18 - 1) \times (- 2) \)
\( = 6 + 17 \times (- 2) \)
\( = 6 - 34 = - 28 \)
Thus, the 18th term is − 28.

Question. The 4th term of an A.P. is 14 and its 12th term 70. What is its first term?
Answer: Let the first term = a
If ‘d’ is the common difference,
Then \( T_4 = a + 3d = 14 \) ...(1)
And \( T_{12} = a + 11d = 70 \) ...(2)
Subtracting (1) from (2),
\( a + 11d - a - 3d = 70 - 14 \)
\( \Rightarrow 8d = 56 \Rightarrow d = \frac{56}{8} = 7 \)
\( \therefore \) From (1), \( a + 3 (7) = 14 \)
\( \Rightarrow a + 21 = 14 \)
\( \Rightarrow a = 14 - 21 = (- 7) \)
Thus, the first term is − 7.

Question. Which term of A.P. 5, 2, − 1, − 4 ..... is − 40?
Answer: Here, \( a = 5, d = 2 - 5 = - 3 \)
Let nth term be − 40
\( \therefore T_n = a + (n - 1) d \)
\( \Rightarrow - 40 = 5 + (n - 1) \times (- 3) \)
\( \Rightarrow n - 1 = \frac{- 40 - 5}{- 3} = \frac{- 45}{- 3} = 15 \)
\( \Rightarrow n = 15 + 1 = 16 \)
i.e., The 16th term of the A.P. is − 40.

Question. What is the sum of all the natural numbers from 1 to 100?
Answer: We have:
1, 2, 3, 4, ....., 100 are in an A.P. such that
\( a = 1 \) and \( l = 100 \)
\( \therefore S_n = \frac{n}{2} [a + l] \)
\( \Rightarrow S_{100} = \frac{100}{2} [1 + 100] = 50 \times 101 = 5050. \)

Question. For an A.P., the 8th term is 17 and the 14th term is 29. Find its common difference.
Answer: Let the common difference = d and first term = a
\( \therefore T_8 = a + 7d = 17 \) ...(1)
\( T_{14} = a + 13d = 29 \) ...(2)
Subtracting (1) from (2), we have:
\( a + 13d - a - 7d = 29 - 17 \)
\( \Rightarrow 6d = 12 \)
\( \Rightarrow d = \frac{12}{6} = 2 \)
\( \therefore \) The required common difference = 2.

Question. If the first and last terms of an A.P. are 10 and − 10. How many terms are there? Given that d = − 1.
Answer: Let the required number of terms is n and
1st term \( a = 10 \)
nth term \( T_n = - 10 \)
Let common difference be d then using,
\( T_n = a + (n - 1) d \), we have:
\( - 10 = 10 + (n - 1) \times (- 1) \)
\( \Rightarrow - 10 = 10 - n + 1 \)
\( \Rightarrow - n + 1 = - 10 - 10 = - 20 \)
\( \Rightarrow - n = - 20 - 1 = - 21 \)
\( \Rightarrow n = 21 \)

Question. The nth term of an A.P. is (3n − 2) find its first term.
Answer: \( \because T_n = 3n - 2 \)
\( \therefore T_1 = 3 (1) - 2 = 3 - 2 = 1 \)
\( \Rightarrow \) First term = 1

Question. The nth term of an A.P. is (2n − 3) find the common difference.
Answer: Here, \( T_n = 2n - 3 \)
\( \therefore T_1 = 2 (1) - 3 = - 1 \)
\( T_2 = 2 (2) - 3 = 1 \)
\( \therefore d = T_2 - T_1 = 1 - (- 1) = 2 \)
Thus the common difference is 2.

Question. If the nth term of an A.P. is (7n − 5). Find its 100th term.
Answer: Here, \( T_n = 7n - 5 \)
\( \therefore T_1 = 7 (1) - 5 = 2 \)
\( T_2 = 7 (2) - 5 = 9 \)
\( \therefore a = 2 \)
and \( d = T_2 - T_1 = 9 - 2 = 7 \)
Now \( T_{100} = 2 + (100 - 1) 7 \) [using \( T_n = a + (n - 1) d \)]
\( = 2 + 99 \times 7 \)
\( = 2 + 693 = 695. \)

Question. Find the sum of first 12 terms of the A.P. 5, 8, 11, 14, ...... .
Answer: Here, \( a = 5, d = 8 - 5 = 3, n = 12 \)
Using \( S_n = \frac{n}{2} [2 (a) + (n - 1) d] \)
we have: \( S_{12} = \frac{12}{2} [2 (5) + (12 - 1) \times 3] \)
\( = 6 [10 + 33] \)
\( = 6 \times 43 = 258 \)

Question. Write the common difference of an A.P. whose nth term is 3n + 5.
Answer: \( T_n = 3n + 5 \)
\( \therefore T_1 = 3 (1) + 5 = 8 \)
\( T_2 = 3 (2) + 5 = 11 \)
\( \Rightarrow d = T_2 - T_1 = 11 - 8 = 3 \)
Thus, the common difference = 3.

Question. Write the value of x for which x + 2, 2x, 2x + 3 are three consecutive terms of an A.P.
Answer: Here, \( T_1 = x + 2, T_2 = 2x, T_3 = 2x + 3 \)
For an A.P., we have:
\( \therefore 2x - (x + 2) = 2x + 3 - 2x \)
\( \Rightarrow 2x - x - 2 = 3 \)
\( \Rightarrow x - 2 = 3 \)
\( \Rightarrow x = 3 + 2 = 5 \)
Thus, x = 5

Question. What is the common difference of an A.P. whose nth term is 3 + 5n?
Answer: \( \because T_n = 3 + 5n \)
\( \therefore T_1 = 3 + 5 (1) = 8 \)
And \( T_2 = 3 + 5 (2) = 13 \)
\( \because d = T_2 - T_1 \)
\( \therefore d = 13 - 8 = 5 \)
Thus, common difference = 5.

Question. For what value of k, are the numbers x, (2x + k) and (3x + 6) three consecutive terms of an A.P.?
Answer: Here, \( T_1 = x, T_2 = (2x + k) \) and \( T_3 = (3x + 6) \)
For an A.P., we have
\( T_2 - T_1 = T_3 - T_2 \)
i.e., \( 2x + k - x = 3x + 6 - (2x + k) \)
\( \Rightarrow x + k = 3x + 6 - 2x - k \)
\( \Rightarrow x + k = x + 6 - k \)
\( \Rightarrow k + k = x + 6 - x \)
\( \Rightarrow 2k = 6 \)
\( \Rightarrow k = \frac{6}{2} = 3 \)

Question. If \( \frac{4}{5} \), a, 2 are three consecutive terms of an A.P., then find the value of a?
Answer: Here, \( T_1 = \frac{4}{5}, T_2 = a, T_3 = 2 \)
\( \because \) For an A.P., \( T_2 - T_1 = T_3 - T_2 \)
\( \therefore a - \frac{4}{5} = 2 - a \)
\( \Rightarrow a + a = 2 + \frac{4}{5} \)
\( \Rightarrow 2a = \frac{14}{5} \)
\( \Rightarrow a = \frac{14}{5} \times \frac{1}{2} = \frac{7}{5} \)
Thus, \( a = \frac{7}{5} \)

Question. For what value of p are 2p − 1, 7 and 3p three consecutive terms of an A.P.?
Answer: Here, \( T_1 = 2p - 1, T_2 = 7, T_3 = 3p \)
\( \because \) For an A.P., we have:
\( T_2 - T_1 = T_3 - T_2 \)
\( \Rightarrow 7 - (2p - 1) = 3p - 7 \)
\( \Rightarrow 7 - 2p + 1 = 3p - 7 \)
\( \Rightarrow - 2p - 3p = - 7 - 1 - 7 \)
\( \Rightarrow - 5p = - 15 \)
\( \Rightarrow p = \frac{- 15}{- 5} = 3 \)
Thus, p = 3

Question. For what value of p are 2p + 1, 13 and 5p − 3 three consecutive terms of an A.P.?
Answer: Here, \( T_1 = 2p + 1, T_2 = 13, T_3 = 5p - 3 \)
For an A.P., we have:
\( T_2 - T_1 = T_3 - T_2 \)
\( \Rightarrow 13 - (2p + 1) = 5p - 3 - 13 \)
\( \Rightarrow 13 - 2p - 1 = 5p - 16 \)
\( \Rightarrow - 2p + 12 = 5p - 16 \)
\( \Rightarrow - 2p - 5p = - 16 - 12 = - 28 \)
\( \Rightarrow - 7p = - 28 \)
\( \Rightarrow p = \frac{- 28}{- 7} = \frac{28}{7} = 4 \)
\( \therefore p = 4 \)

Question. The nth term of an A.P. is 7 − 4n. Find its common difference.
Answer: \( \because T_n = 7 - 4n \)
\( \therefore T_1 = 7 - 4 (1) = 3 \)
\( T_2 = 7 - 4 (2) = - 1 \)
\( \therefore d = T_2 - T_1 = (- 1) - 3 = - 4 \)
Thus, common difference = − 4

Question. The nth term of an A.P. is 6n + 2. Find the common difference.
Answer: Here, \( T_n = 6n + 2 \)
\( \therefore T_1 = 6 (1) + 2 = 8 \)
\( T_2 = 6 (2) + 2 = 14 \)
\( \Rightarrow d = T_2 - T_1 = 14 - 8 = 6 \)
\( \therefore \) Common difference = 6.

Question. Write the next term of the A.P. \( \sqrt{8}, \sqrt{18}, \sqrt{32} \), .....
Answer: Here, \( T_1 = \sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2} \)
\( T_2 = \sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2} \)
\( T_3 = \sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2} \)
\( \therefore a = 2\sqrt{2} \)
Now, \( d = T_2 - T_1 = 3\sqrt{2} - 2\sqrt{2} = \sqrt{2}(3 - 2) = \sqrt{2} \)
\( \therefore T_4 = a + 3d = 2\sqrt{2} + 3(\sqrt{2}) = 2\sqrt{2} + 3\sqrt{2} = \sqrt{2}(2 + 3) = 5\sqrt{2} \text{ or } \sqrt{50} \)
Thus, the next term of the A.P. is \( 5\sqrt{2} \) or \( \sqrt{50} \).

Question. The first term of an A.P. is p and its common difference is q. Find the 10th term.
Answer: Here, \( a = p \) and \( d = q \)
\( \because T_n = a + (n - 1) d \)
\( \therefore T_{10} = p + (10 - 1) q = p + 9q \)
Thus, the 10th term is p + 9q.

Question. Find the next term of the A.P. \( \sqrt{2}, \sqrt{8}, \sqrt{18} \), .....
Answer: Here, \( T_1 = \sqrt{2} \Rightarrow a = \sqrt{2} \)
\( T_2 = \sqrt{8} = 2\sqrt{2} \)
\( T_3 = \sqrt{18} = 3\sqrt{2} \)
Now, \( d = T_2 - T_1 = 2\sqrt{2} - \sqrt{2} = \sqrt{2} \)
Now, using \( T_n = a + (n - 1) \times d \), we have
\( T_4 = a + 3d = \sqrt{2} + 3(\sqrt{2}) = \sqrt{2}[1 + 3] = 4\sqrt{2} = \sqrt{16 \times 2} = \sqrt{32} \)
Thus, the next term = \( \sqrt{32} \).

Question. Which term of the A.P.: 21, 18, 15, ..... is zero?
Answer: Here, \( a = 21, d = 18 - 21 = - 3 \)
Since \( T_n = a + (n - 1) d \)
\( \Rightarrow 0 = 21 + (n - 1) \times (- 3) \)
\( \Rightarrow - 3 (n - 1) = - 21 \)
\( \Rightarrow (n - 1) = \frac{- 21}{- 3} = 7 \)
\( \Rightarrow n = 7 + 1 = 8 \)
Thus, the 8th term of this A.P. will be 0.

Question. Which term of the A.P.: 14, 11, 8, ..... is − 1?
Answer: Here, \( a = 14, d = 11 - 14 = - 3 \)
Let the nth term be (− 1)
\( \therefore \) Using \( T_n = a + (n - 1) d \), we get
\( - 1 = 14 + (n - 1) \times (- 3) \)
\( \Rightarrow - 1 - 14 = - 3 (n - 1) \)
\( \Rightarrow - 15 = - 3 (n - 1) \)
\( \therefore n - 1 = \frac{- 15}{- 3} = 5 \)
\( \Rightarrow n = 5 + 1 = 6 \)
Thus, −1 is the 6th term of the A.P.

Question. The value of the middlemost term (s) of the AP : –11, –7, –3, ...49.
Answer: \( \because a = –11, a_n = 49 \) and \( d = (–7) – (–11) = 4 \)
\( \therefore a_n = a + (n – 1)d \)
\( \Rightarrow 49 = –11 + (n – 1) \times 4 \Rightarrow n = 16 \)
Since, n is an even number
\( \therefore \) There will be two middle terms, which are:
\( \frac{16}{2} \)th and \( (\frac{16}{2} + 1) \)th
or 8th and 9th
Now, \( a_8 = a + (8 – 1)d = –11 + 7 \times 4 = 17 \)
\( a_9 = a + (9 – 1)d = –11 + 8 \times 4 = 21 \)
Thus, the values of the two middlemost terms are : 17 and 21.