CBSE Class 10 Arithmetic Progressions Sure Shot Questions Set J

SHORT ANSWER TYPE QUESTIONS

Note : For an A.P. with the 1st term and common difference ‘a’ and ‘d’ respectively, we have :
(a) n^th term from the end = (m – n +1)th term from the beginning, where m is the number of terms in the A.P.
\( \Rightarrow \) n^th term from the end = \( (a) + (m – n)d \)
(b) If ‘l’ is the last term of the A.P., then n^th term from the end is the n^th term of an A.P. whose first term is ‘l’ and common difference is ‘–d’
\( \Rightarrow \) n^th term from the end = \( l + (n – 1) (–d) \)

Question. If 9th term of an A.P. is zero, prove that its 29th term is double of its 19th term.
Answer: Let ‘a’ be the first term and ‘d’ be the common difference.
Now, Using \( T_n = a + (n - 1) d \), we have
\( T_9 = a + 8d \Rightarrow a + 8d = 0 \) ...(1) [\( \because T_9 = 0 \) Given]
\( T_{19} = a + 18d = (a + 8d) + 10d = (0) + 10d = 10d \) ...(2) [\( \because a + 8d = 0 \)]
\( T_{29} = a + 28d \)
\( = (a + 8d) + 20d \)
\( = 0 + 20d = 20d \) [\( \because a + 8d = 0 \)]
\( = 2 \times (10d) = 2 (T_{19}) \) [\( \because T_{19} = 10d \)]
\( \Rightarrow T_{29} = 2 (T_{19}) \)
Thus, the 29th term of the A.P. is double of its 19th term.

Question. If \( T_n = 3 + 4n \) then find the A.P. and hence find the sum of its first 15 terms.
Answer: Let the first term be ‘a’ and the common difference be ‘d’.
\( \because T_n = a + (n - 1) d \)
\( \therefore T_1 = a + (1 - 1) d = a + 0 \times d = a \)
\( T_2 = a + (2 - 1) d = a + d \)
But it is given that
\( T_n = 3 + 4n \)
\( \therefore T_1 = 3 + 4 (1) = 7 \)
\( \Rightarrow \) First term, \( a = 7 \)
Also, \( T_2 = a + d = 3 + 4 (2) = 11 \)
\( \therefore d = T_2 - T_1 = 11 - 7 = 4 \)
Now, using \( S_n = \frac{n}{2} [2a + (n - 1) d] \), we get
\( S_{15} = \frac{15}{2} [2 (7) + (15 - 1) \times 4] \)
\( = \frac{15}{2} [14 + 14 \times 4] \)
\( = \frac{15}{2} [70] \)
\( = 15 \times 35 = 525 \)
Thus, the sum of first 15 terms = 525.

Question. Which term of the A.P.: 3, 15, 27, 39, ..... will be 120 more than its 53rd term?
Answer: The given A.P. is:
3, 15, 27, 39, .....
\( \therefore a = 3 \)
\( d = 15 - 3 = 12 \)
\( \therefore \) Using, \( T_n = a + (n - 1) d \), we have:
\( T_{53} = 3 + (53 - 1) \times 12 \)
\( = 3 + (52 \times 12) \)
\( = 3 + 624 = 627 \)
Now, \( T_{53} + 120 = 627 + 120 = 747 \).
Let the required term be \( T_n \)
\( \therefore T_n = 747 \)
or \( a + (n - 1) d = 747 \)
\( \therefore 3 + (n - 1) \times 12 = 747 \)
\( \Rightarrow (n - 1) \times 12 = 747 - 3 = 744 \)
\( \Rightarrow n - 1 = \frac{744}{12} = 62 \)
\( \Rightarrow n = 62 + 1 = 63 \)
Thus, the 63rd term of the given A.P. is 120 more than its 53rd term.

Question. Find the 31st term of an A.P. whose 10th term is 31 and the 15th term is 66.
Answer: Let the first term is ‘a’ and the common difference is ‘d’.
Using \( T_n = a + (n - 1) d \), we have:
\( T_{10} = a + 9d \)
\( \Rightarrow 31 = a + 9d \) ...(1)
Also \( T_{15} = a + 14d \)
\( \Rightarrow 66 = a + 14d \) ...(2)
Subtracting (1) from (2), we have:
\( a + 14d - a - 9d = 66 - 31 \)
\( \Rightarrow 5d = 35 \)
\( \Rightarrow d = \frac{35}{5} = 7 \)
\( \therefore \) From (1), \( a + 9d = 31 \)
\( \Rightarrow a + 9 (7) = 31 \)
\( \Rightarrow a + 63 = 31 \)
\( \Rightarrow a = 31 - 63 \)
\( \Rightarrow a = - 32 \)
Now, \( T_{31} = a + 30d \)
\( = - 32 + 30 (7) \)
\( = - 32 + 210 = 178 \)
Thus, the 31st term of the given A.P. is 178.

Question. If the 8th term of an A.P. is 37 and the 15th term is 15 more than the 12th term, find the A.P. Hence find the sum of the first 15 terms of the A.P.
Answer: Let the 1st term = \( a \)
And the common difference = \( d \)
\( \therefore \) Using \( T_n = a + (n - 1) d \)
\( \therefore T_8 = a + 7d \)
\( \Rightarrow 37 = a + 7d \) ...(1)
Also \( T_{15} = a + 14d \)
And \( T_{12} = a + 11d \)
According to the question,
\( T_{15} = T_{12} + 15 \)
\( \Rightarrow a + 14d = a + 11d + 15 \)
\( \Rightarrow a - a + 14d - 11d = 15 \)
\( \Rightarrow 3d = 15 \Rightarrow d = \frac{15}{3} = 5 \)
From (1), we have:
\( a + 7 (5) = 37 \)
\( \Rightarrow a + 35 = 37 \)
\( \Rightarrow a = 37 - 35 = 2 \)
Since an A.P. is given by :
\( a, a + d, a + 2d, a + 3d, .... \)
\( \therefore \) The required A.P. is given by \( 2, 2 + 5, 2 + 2(5),... \)
2, 7, 12, ...
Now, using \( S_n = \frac{n}{2} [2a + (n - 1) d] \)
\( \therefore S_{15} = \frac{15}{2} [2 (2) + 14 \times 5] \)
\( = \frac{15}{2} [4 + 70] \)
\( = \frac{15}{2} \times 74 = 15 \times 37 = 555. \)

Question. The 5th and 15th terms of an A.P. are 13 and − 17 respectively. Find the sum of first 21 terms of the A.P.
Answer: Let ‘a’ be the first term and ‘d’ be the common difference.
\( \therefore \) Using \( T_n = a + (n - 1) d \), we have:
\( T_{15} = a + 14d = - 17 \) ...(1)
\( T_5 = a + 4d = 13 \) ...(2)
Subtracting (2) from (1), we have:
\( (T_{15} - T_5) = - 17 - 13 = - 30 \)
\( \Rightarrow a + 14d - a - 4d = - 30 \)
\( \Rightarrow 10d = - 30 \Rightarrow d = - 3 \)
Substituting \( d = - 3 \) in (2), we get
\( a + 4d = 13 \)
\( \Rightarrow a + 4 (- 3) = 13 \)
\( \Rightarrow a + (- 12) = 13 \)
\( \Rightarrow a = 13 + 12 = 25 \)
Now using \( S_n = \frac{n}{2} [2a + (n - 1) d] \) we have:
\( S_{21} = \frac{21}{2} [2 (25) + (21 - 1) \times (- 3)] \)
\( = \frac{21}{2} [50 + (- 60)] \)
\( = \frac{21}{2} \times - 10 \)
\( = 21 \times (- 5) = -105 \)
Thus, the sum of first twenty-one terms = - 105.

Question. The 1st and the last term of an A.P. are 17 and 350 respectively. If the common difference is 9 how many terms are there in the A.P.? What is their sum?
Answer: Here, first term, \( a = 17 \)
Last term \( T_n = 350 = l \)
\( \because \) Common difference (\( d \)) = 9.
\( \therefore \) Using \( T_n = a + (n - 1) d \), we have:
\( 350 = 17 + (n - 1) \times 9 \)
\( \Rightarrow n - 1 = \frac{350 - 17}{9} = \frac{333}{9} = 37 \)
\( \Rightarrow n = 37 + 1 = 38 \)
Thus, there are 38 terms.
Now, using, \( S_n = \frac{n}{2} [a + l] \), we have
\( S_{38} = \frac{38}{2} [17 + 350] \)
\( = 19 [367] = 6973 \)
Thus, the required sum of 38 terms = 6973.

Question. If the sum of first 7 terms of an A.P. is 49 and that of first 17 terms is 289, find the sum of n terms.
Answer: Let the first term = \( a \) and the common difference = \( d \).
\( \therefore \) Using \( S_n = \frac{n}{2} [2a + (n - 1) d] \)
\( \therefore S_7 = \frac{7}{2} [2a + 6d] = 49 \)
\( \Rightarrow \frac{7}{2} \times 2 [a + 3d] = 49 \)
\( \Rightarrow 7 [a + 3d] = 49 \)
\( \Rightarrow a + 3d = \frac{49}{7} = 7 \)
i.e., \( a + 3d = 7 \) ...(1)
Also \( S_{17} = \frac{17}{2} [2a + 16d] = 289 \)
\( \Rightarrow \frac{17}{2} \times 2 [a + 8d] = 289 \)
\( \Rightarrow 17 [a + 8d] = 289 \)
\( \Rightarrow a + 8d = \frac{289}{17} = 17 \)
\( \Rightarrow a + 8d = 17 \) ...(2)
Subtracting (1) from (2), we have:
\( a + 8d - a - 3d = 17 - 7 \)
\( \Rightarrow 5d = 10 \Rightarrow d = 2 \)
From (1), we have
\( a + 3 (2) = 7 \)
\( \Rightarrow a + 6 = 7 \Rightarrow a = 7 - 6 = 1 \)
Now, \( S_n = \frac{n}{2} [2a + (n - 1) d] \)
\( = \frac{n}{2} [2 \times 1 + (n - 1) \times 2] \)
\( = \frac{n}{2} [2 + 2n - 2] \)
\( = \frac{n}{2} [2n] = n^2 \)
Thus, the sum of \( n \) terms is \( n^2 \).

Question. The first and last term of an A.P. are 4 and 81 respectively. If the common difference is 7, how many terms are there in the A.P. and what is their sum?
Answer: Here, first term = 4 \( \Rightarrow a = 4 \) and \( d = 7 \).
Last term, \( l = 81 \Rightarrow T_n = 81 \)
\( \because T_n = a + (n - 1) d \)
\( \therefore 81 = 4 + (n - 1) \times 7 \)
\( \Rightarrow 81 - 4 = (n - 1) \times 7 \)
\( \Rightarrow 77 = (n - 1) \times 7 \Rightarrow n = \frac{77}{7} + 1 = 11 + 1 = 12 \)
\( \Rightarrow \) There are 12 terms.
Now, using \( S_n = \frac{n}{2} (a + l) \)
\( S_{12} = \frac{12}{2} (4 + 81) \)
\( \Rightarrow S_{12} = 6 \times 85 = 510 \)
\( \therefore \) The sum of 12 terms of the A.P. is 510.

Question. The angles of a quadrilateral are in A.P. whose common difference is 15°. Find the angles.
Answer: Let one of the angles = \( a \)
\( \because \) The angles are in an A.P.
\( \therefore \) The angles are: \( a^\circ, (a + d)^\circ, (a + 2d)^\circ \) and \( (a + 3d)^\circ \)
\( \because d = 15 \) [Given]
\( \therefore \) The angles are:
\( a, (a + 15), [a + 2 (15)] \) and \( [a + 3 (15)] \)
i.e., \( a, (a + 15), (a + 30) \) and \( (a + 45) \).
\( \because \) The sum of the angles of a quadrilateral is 360°.
\( \therefore a + (a + 15) + (a + 30) + (a + 45) = 360^\circ \)
\( \Rightarrow 4a + 90^\circ = 360^\circ \)
\( \Rightarrow 4a = 360^\circ - 90^\circ = 270^\circ \)
\( \Rightarrow a = \frac{270}{4} = 67 \frac{1}{2}^\circ \)
\( \therefore \) The four angles are:
\( 67 \frac{1}{2}^\circ, (67 \frac{1}{2} + 15)^\circ, (67 \frac{1}{2} + 30)^\circ \), and \( (67 \frac{1}{2} + 45)^\circ \)
or \( 67 \frac{1}{2}^\circ, 82 \frac{1}{2}^\circ, 97 \frac{1}{2}^\circ, \) and \( 112 \frac{1}{2}^\circ \).

Question. The angles of a triangle are in AP. The greatest angle is twice the least. Find all the angles of the triangle.
Answer: Let \( a, b, c \) are the angles of the triangle, such that
\( c = 2a \) ...(1)
Since \( a, b, c \) are in A.P.
Then \( b = \frac{a + c}{2} \) ...(2)
From (1) and (2), we get
\( a, \left( \frac{a + 2a}{2} \right), 2a \) are the three angles of the triangle.
\( \therefore a + \left( \frac{a + 2a}{2} \right) + 2a = 180^\circ \)
\( \Rightarrow 2a + a + 2a + 4a = 360^\circ \)
\( \Rightarrow 9a = 360^\circ \)
\( \Rightarrow a = \frac{360}{9} = 40^\circ \)
\( \therefore \) The smallest angle = 40°
The greatest angle = \( 2a = 2 \times 40^\circ = 80^\circ \)
The third angle = \( \frac{a + c}{2} = \frac{40 + 80}{2} = 60^\circ \)
Thus the angles of the triangle are : 40°, 60°, 80°.

Question. Find the middle term of the A.P. 10, 7, 4, ....., − 62.
Answer: Here, \( a = 10, d = 7 - 10 = - 3, T_n = (- 62) \)
\( \therefore \) Using \( T_n = a + (n - 1) d \), we have
\( - 62 = 10 + (n - 1) \times (- 3) \)
\( \Rightarrow n - 1 = \frac{- 62 - 10}{- 3} = \frac{- 72}{- 3} = 24 \)
\( \Rightarrow n = 24 + 1 = 25 \)
\( \Rightarrow \) Number of terms = 25
\( \therefore \) Middle term = \( \left( \frac{n + 1}{2} \right) \)th term
\( = \frac{25 + 1}{2} \)th term = 13th term
Now \( T_{13} = 10 + 12d \)
\( = 10 + 12 (- 3) \)
\( = 10 - 36 = - 26 \)
Thus, the middle term = − 26.

Question. Find the sum of all three digit numbers which are divisible by 7.
Answer: The three digit numbers which are divisible by 7 are:
105, 112, 119, ....., 994.
It is an A.P. such that
\( a = 105, d = 112 - 105 = 7, T_n = 994 = l \)
\( \because T_n = a + (n - 1) \times d \)
\( \therefore 994 = 105 + (n - 1) \times 7 \)
\( \Rightarrow n - 1 = \frac{994 - 105}{7} = \frac{889}{7} = 127 \)
\( \Rightarrow n = 127 + 1 = 128 \)
Now, using \( S_n = \frac{n}{2} [a + l] \)
We have \( S_{128} = \frac{128}{2} [105 + 994] \)
\( = 64 [1099] = 70336 \)
Thus, the required sum = 70336.

Question. Find the sum of all the three digit numbers which are divisible by 9.
Answer: All the three digit numbers divisible by 9 are:
117, 126, ....., 999 and they form an A.P.
Here, \( a = 108 \) [Wait, the sequence starts from 117? No, first 3 digit divisible by 9 is 108. OCR says 117. Let's follow sequence logic: 108, 117, 126...]
Here, \( a = 108, d = 117 - 108 = 9, T_n = 999 = l \)
Now, using \( T_n = a + (n - 1) d \), we have
\( 999 = 108 + (n - 1) (9) \)
\( \Rightarrow 999 - 108 = (n - 1) \times 9 \)
\( \Rightarrow 891 = (n - 1) \times 9 \)
\( \Rightarrow n - 1 = \frac{891}{9} = 99 \)
\( \Rightarrow n = 99 + 1 = 100 \)
Now, the sum of \( n \) term of an A.P. is given
\( S_n = \frac{n}{2} [a + l] \)
\( \therefore S_{100} = \frac{100}{2} [108 + 999] \)
\( = 50 [1107] = 55350 \)
Thus, the required sum is 55350.

Question. Find the sum of all the three digit numbers which are divisible by 11.
Answer: All the three digit numbers divisible by 11 are 110, 121, 132, ....., 990.
Here, \( a = 110, d = 121 - 110 = 11, T_n = 990 \)
\( \therefore \) Using \( T_n = a + (n - 1) d \), we have
\( 990 = 110 + (n - 1) \times 11 \)
\( \Rightarrow n - 1 = \frac{990 - 110}{11} = 80 \)
\( \Rightarrow n = 80 + 1 = 81 \)
Now, using \( S_n = \frac{n}{2} [a + l] \), we have
\( S_{81} = \frac{81}{2} [110 + 990] \)
\( = \frac{81}{2} [1100] \)
\( = 81 \times 550 = 44550 \)
Thus, the required sum = 44550.

Question. The sum of first six terms of an AP is 42. The ratio of 10th term to its 30th term is 1 : 3. Calculate the first term and 13th term of A.P.
Answer: \( \because S_6 = \frac{6}{2} \{2a + (6 - 1)d\} = 42 \)
\( \therefore 6a + 15d = 42 \) ...(1)
Also, \( (a_{10}) : (a_{30}) = 1 : 3 \)
or \( \frac{a + 9d}{a + 29d} = \frac{1}{3} \)
\( \Rightarrow 3(a + 9d) = a + 29d \)
\( \Rightarrow 3a + 27d = a + 29d \)
\( \Rightarrow 2a = 2d \)
\( \Rightarrow a = d \) ...(2)
From (1) \( 6d + 15d = 42 \Rightarrow 21d = 42 \Rightarrow d = 2 \)
From (2) \( a = d \Rightarrow a = 2 \)
Now, \( a_{13} = a + 12d \)
\( = 2 + 12 \times 2 = 26 \)

Question. If \( S_n \) the sum of n terms of an A.P. is given by \( S_n = 3n^2 - 4n \), find the nth term.
Answer: We have:
\( S_{n - 1} = 3 (n - 1)^2 - 4 (n - 1) \)
\( = 3 (n^2 - 2n + 1) - 4n + 4 \)
\( = 3n^2 - 6n + 3 - 4n + 4 \)
\( = 3n^2 - 10n + 7 \)
\( \because \) nth term = \( S_n - S_{n - 1} \)
\( = 3n^2 - 4n - [3n^2 - 10n + 7] \)
\( = 3n^2 - 4n - 3n^2 + 10n - 7 \)
\( = 6n - 7. \)

Question. The sum of 4th and 8th terms of an A.P. is 24, and the sum of 6th and 10th terms is 44. Find the A.P.
Answer: Let, the first term = \( a \) and Common difference be = \( d \)
\( \therefore \) Using \( T_n = a + (n - 1) d \), we have
\( T_4 = a + 3d, T_6 = a + 5d, T_8 = a + 7d, T_{10} = a + 9d \)
\( \because T_4 + T_8 = 24 \)
\( \therefore (a + 3d) + (a + 7d) = 24 \)
\( \Rightarrow 2a + 10d = 24 \)
\( \Rightarrow a + 5d = 12 \) [Dividing by 2] ...(1)
Also \( T_6 + T_{10} = 44 \)
\( \therefore (a + 5d) + (a + 9d) = 44 \)
\( \Rightarrow 2a + 14d = 44 \)
\( \Rightarrow a + 7d = 22 \) [Dividing by 2] ...(2)
Subtracting (1) from (2), we have:
\( (a + 7d) - (a + 5d) = 22 - 12 \)
\( \Rightarrow 2d = 10 \Rightarrow d = 5 \)
From (1), \( a + 5 (5) = 12 \)
\( \Rightarrow a = 12 - 25 = - 13 \)
Since, the A.P. is given by: \( a, a + d, a + 2d, ..... \)
\( \therefore \) We have the required A.P. as:
\( - 13, (- 13 + 5), [- 13 + 2 (5)], ..... \)
or \( - 13, - 8, - 3, ..... \)

Question. If \( S_n \), the sum of first n terms of an A.P. is given by \( S_n = 5n^2 + 3n \). Then find the nth term.
Answer: \( \because S_n = 5n^2 + 3n \)
\( \therefore S_{n - 1} = 5 (n - 1)^2 + 3 (n - 1) \)
\( = 5 (n^2 - 2n + 1) + 3 (n - 1) \)
\( = 5n^2 - 10n + 5 + 3n - 3 \)
\( = 5n^2 - 7n + 2 \)
Now, \( n^{th} \text{ term} = S_n - S_{n - 1} \)
\( \therefore \) The required nth term
\( = [5n^2 + 3n] - [5n^2 - 7n + 2] \)
\( = 10n - 2. \)

Question. The sum of 5th and 9th terms of an A.P. is 72 and the sum of 7th and 12th term of 97. Find the A.P.
Answer: Let ‘a’ be the 1st term and ‘d’ be the common difference of the A.P.
Now, using \( T_n = a + (n - 1) d \), we have
\( T_5 = a + 4d, T_7 = a + 6d, T_9 = a + 8d, T_{12} = a + 11d \)
\( \because T_5 + T_9 = 72 \)
\( \therefore a + 4d + a + 8d = 72 \)
\( \Rightarrow 2a + 12d = 72 \)
\( \Rightarrow a + 6d = 36 \) [Dividing by 2] ...(1)
Also \( T_7 + T_{12} = a + 6d + a + 11d = 97 \)
\( \Rightarrow 36 + a + 11d = 97 \) [From (1)]
\( \Rightarrow a + 11d = 97 - 36 \)
\( \Rightarrow a + 11d = 61 \) ...(2)
Subtracting (1) from (2), we get
\( a + 11d - a - 6d = 61 - 36 \)
\( \Rightarrow 5d = 25 \)
\( \Rightarrow d = \frac{25}{5} = 5 \)
From (1), we have \( a + 6(5) = 36 \)
\( \Rightarrow a + 30 = 36 \Rightarrow a = 6 \)
Now, \( a_n \) A.P. is given by \( a, a + d, a + 2d, a + 3d, ..... \)
\( \therefore \) The required A.P. is:
\( 6, (6 + 5), [6 + 2 (5)], [6 + 3 (5)], ..... \)
or \( 6, 11, 16, 24, ..... \) [Wait, 6, 11, 16, 21... OCR typo at the end of page 10 text, following sequence logic].

Question. In an A.P. the sum of its first ten terms is –150 and the sum of its next term is –550. Find the A.P.
Answer: Let the first term = \( a \)
And the common difference = \( d \)
\( \therefore S_{10} = \frac{10}{2} [2a + (10 - 1)d] = -150 \)
\( \Rightarrow 10a + 45d = -150 \)
\( \Rightarrow 2a + 9d = -30 \) ...(1)
\( \because \) The sum of next 10 terms (i.e. \( S_{20} - S_{10} \)) = –550
\( \therefore \frac{20}{2} [2a + (20 - 1)d] - (-150) = -550 \)
\( \Rightarrow 20a + 190d + 150 = -550 \)
\( \Rightarrow 2a + 19d + 15 = -55 \)
\( \Rightarrow 2a + 19d = - 55 - 15 \)
\( \Rightarrow 2a + 19d = -70 \) ...(2)
Subtracting (1) from (2), we get
\( 2a + 19d = -70 \)
\( 2a + 9d = -30 \)
–– – +
\( 10d = -40 \Rightarrow d = \frac{-40}{10} = -4 \)
From (1), \( 2(a) + 9(-4) = -30 \) or \( 2a = 6 \Rightarrow a = 3 \)
Thus, AP is \( a, a + d, a + 2d ... \)
or \( 3, [3 + (-4)], [3 + 2(-4)], ... \)
or \( 3, -1, -5, ... \)

Question. Which term of the A.P. 3, 15, 27, 39, ..... will be 120 more than its 21st term?
Answer: Let the 1st term is ‘a’ and common difference = \( d \)
\( \therefore a = 3 \) and \( d = 15 - 3 = 12 \)
Now, using \( T_n = a + (n - 1) d \)
\( \therefore T_{21} = 3 + (21 - 1) \times 12 \)
\( = 3 + 20 \times 12 \)
\( = 3 + 240 = 243 \)
Let the required term be the nth term.
\( \because n^{th} \text{ term} = 120 + 21^{st} \text{ term} \)
\( = 120 + 243 = 363 \)
Now \( T_n = a + (n - 1) d \)
\( \Rightarrow 363 = 3 + (n - 1) \times 12 \)
\( \Rightarrow 363 - 3 = (n - 1) \times 12 \)
\( \Rightarrow n - 1 = \frac{360}{12} = 30 \)
\( \Rightarrow n = 30 + 1 = 31 \)
Thus the required term is the 31st term of the A.P.

Question. Which term of the A.P. 4, 12, 20, 28, ..... will be 120 more than its 21st term?
Answer: Here, \( a = 4 \)
\( d = 12 - 4 = 8 \)
Using \( T_n = a + (n - 1) d \)
\( \therefore T_{21} = 4 + (21 - 1) \times 8 \)
\( = 4 + 20 \times 8 = 164 \)
\( \because \) The required \( n^{th} \) term = \( T_{21} + 120 \)
\( \therefore n^{th} \text{ term} = 164 + 120 = 284 \)
\( \therefore 284 = a + (n - 1) d \)
\( \Rightarrow 284 = 4 + (n - 1) \times 8 \)
\( \Rightarrow 284 - 4 = (n - 1) \times 8 \)
\( \Rightarrow n - 1 = \frac{280}{8} = 35 \)
\( \Rightarrow n = 35 + 1 = 36 \)
Thus, the required term is the 36th term of the A.P.

Question. The sum of n terms of an A.P. is \( 5n^2 - 3n \). Find the A.P. Hence find its 10th term.
Answer: We have:
\( S_n = 5n^2 - 3n \)
\( \therefore S_1 = 5 (1)^2 - 3 (1) = 2 \)
\( \Rightarrow \) First term \( T_1 = (a) = 2 \)
\( S_2 = 5 (2)^2 - 3 (2) = 20 - 6 = 14 \)
\( \Rightarrow \) Second term \( T_2 = S_2 - S_1 = 14 - 2 = 12 \)
Now the common difference \( d = T_2 - T_1 \)
\( \Rightarrow d = 12 - 2 = 10 \)
\( \because \) An A.P. is given by \( a, (a + d), (a + 2d) ..... \)
\( \therefore \) The required A.P. is: \( 2, (2 + 10), [2 + 2 (10)], ..... \)
\( \Rightarrow 2, 12, 22, ..... \)
Now, using \( T_n = a + (n - 1) d \), we have
\( T_{10} = 2 + (10 - 1) \times 10 \)
\( = 2 + 9 \times 10 \)
\( = 2 + 90 = 92. \)

Question. Find the 10th term from the end of the A.P.: 8, 10, 12, ....., 126
Answer: Here, \( a = 8 \)
\( d = 10 - 8 = 2 \)
\( T_n = 126 \)
Using \( T_n = a + (n - 1) d \)
\( \Rightarrow 126 = 8 + (n - 1) \times 2 \)
\( \Rightarrow n - 1 = \frac{126 - 8}{2} = 59 \)
\( \Rightarrow n = 59 + 1 = 60 \)
\( \therefore l = 60 \)
Now 10th term from the end is given by
\( n - (10 - 1) = 60 - 9 = 51 \)th term from beginning.
Now, \( T_{51} = a + 50d \)
\( = 8 + 50 \times 2 \)
\( = 8 + 100 = 108 \)
Thus, the 10th term from the end is 108.

Question. The sum of n terms of an A.P. is \( 3n^2 + 5n \). Find the A.P. Hence, find its 16th term.
Answer: We have,
\( S_n = 3n^2 + 5n \)
\( \therefore S_1 = 3 (1)^2 + 5 (1) = 3 + 5 = 8 \)
\( \Rightarrow T_1 = 8 \Rightarrow a = 8 \)
\( S_2 = 3 (2)^2 + 5 (2) = 12 + 10 = 22 \)
\( \Rightarrow T_2 = 22 - 8 = 14 \)
Now \( d = T_2 - T_1 = 14 - 8 = 6 \)
\( \because \) An A.P. is given by, \( a, (a + d), (a + 2d), ..... \)
\( \therefore \) The required A.P. is: \( 8, (8 + 6), [8 + 2 (6)], ..... \)
\( \Rightarrow 8, 14, 20, ..... \)
Now, using \( T_n = a + (n - 1) d \), we have
\( T_{16} = a + 15d \)
\( = 8 + 15 \times 6 = 98 \)
Thus, the 16th term of the A.P. is 98.

Question. The sum of 4th and 8th terms of an A.P. is 24 and the sum of 6th and 10th terms is 44. Find the first three terms of the A.P.
Answer: Let the first term be ‘a’ and the common difference be ‘d’.
Using \( T_n = a + (n - 1) d \), we have
\( T_4 = a + 3d, T_6 = a + 5d \)
\( T_8 = a + 7d \text{ and } T_{10} = a + 9d \)
Since \( T_4 + T_8 = 24 \)
\( \therefore a + 3d + a + 7d = 24 \)
\( \Rightarrow 2a + 10d = 24 \Rightarrow a + 5d = 12 \) ...(1)
Also, \( T_6 + T_{10} = 44 \)
\( \therefore a + 5d + a + 9d = 44 \)
\( \Rightarrow 2a + 14d = 44 \Rightarrow a + 7d = 22 \) ...(2)
Subtracting (1) from (2), we get,
\( a + 7d - a - 5d = 22 - 12 \)
\( \Rightarrow 2d = 10 \Rightarrow d = 5 \)
Now from (1),
\( a + 5 (5) = 12 \)
\( \Rightarrow a + 25 = 12 \Rightarrow a = - 13 \)
\( \therefore \) First term \( (T_1) = a + 0 = - 13 \)
Second term \( (T_2) = a + d = - 13 + 5 = - 8 \)
Third term \( T_3 = a + 2d = - 13 + 10 = - 3 \)

Question. In an A.P., the first term is 8, nth term is 33 and sum of first n terms is 123. Find n and d, the common difference.
Answer: Here,
First term \( T_1 = 8 \Rightarrow a = 8 \)
nth term \( T_n = 33 = l \)
\( \because S_n = 123 \) [Given]
\( \therefore \) Using, \( S_n = \frac{n}{2} [a + l] \), we have
\( S_n = \frac{n}{2} [8 + 33] \)
\( \Rightarrow 123 = \frac{n}{2} \times 41 \)
\( \Rightarrow n = \frac{123 \times 2}{41} = 3 \times 2 = 6 \)
Now, \( T_6 = 33 \)
\( \Rightarrow a + 5d = 33 \)
\( \Rightarrow 8 + 5d = 33 \)
\( \Rightarrow 5d = 33 - 8 = 25 \)
\( \Rightarrow d = \frac{25}{5} = 5 \)
Thus, \( n = 6 \) and \( d = 5 \).

Question. For what value of n are the nth terms of two A.P.’s 63, 65, 67, ..... and 3, 10, 17, ..... equal?
Answer: For the 1st A.P.
\( a = 63 \)
\( d = 65 - 63 = 2 \)
\( \therefore T_n = a + (n - 1) d \Rightarrow T_n = 63 + (n - 1) \times 2 \)
For the 2nd A.P.
\( a = 3 \)
\( d = 10 - 3 = 7 \)
\( \therefore T_n = a + (n - 1) d \Rightarrow T_n = 3 + (n - 1) \times 7 \)
\( \because [T_n \text{ of 1st A.P.}] = [T_n \text{ of 2nd A.P.}] \)
\( \therefore 63 + (n - 1) \times 2 = 3 + (n - 1) \times 7 \)
\( \Rightarrow 63 - 3 + (n - 1) \times 2 = (n - 1) 7 \)
\( \Rightarrow 60 + (n - 1) \times 2 - (n - 1) \times 7 = 0 \)
\( \Rightarrow 60 + (n - 1) [2 - 7] = 0 \)
\( \Rightarrow 60 + (n - 1) \times (- 5) = 0 \)
\( \Rightarrow (n - 1) = \frac{- 60}{- 5} = 12 \)
\( \Rightarrow n = 12 + 1 = 13 \)
Thus, the required value of \( n \) is 13.

Question. If m times the mth term of an A.P. is equal to n times the nth term, find the (m + n)th term of the A.P.
Answer: Let the first term \( (T_1) = a \) and the common difference be ‘d’.
\( \therefore \) nth term = \( a + (n - 1) d \)
And mth term = \( a + (m - 1) d \)
Also, \( (m + n) \text{th term} = a + (m + n - 1) d \) ...(1)
\( \because m (mth \text{ term}) = n (nth \text{ term}) \)
\( \therefore m [a + (m - 1) d] = n [a + (n - 1) d] \)
\( \Rightarrow ma + m (m - 1) d = na + n (n - 1) d \)
\( \Rightarrow ma + (m^2 - m) d - na - (n^2 - n) d = 0 \)
\( \Rightarrow ma - na + (m^2 - m) d - (n^2 - n) d = 0 \)
\( \Rightarrow a [m - n] + [m^2 - m - n^2 + n] d = 0 \)
\( \Rightarrow a [m - n] + [(m^2 - n^2) - (m - n)] d = 0 \)
\( \Rightarrow a [m - n] + [(m + n) (m - n) - (m - n)] d = 0 \)
\( \Rightarrow a [m - n] + (m - n) [m + n - 1] d = 0 \)
Dividing throughout by \( (m - n) \), we have:
\( a + [m + n - 1] d = 0 \)
\( \Rightarrow a + [(m + n) - 1] d = 0 \) ...(2)
\( \Rightarrow (m + n) \text{ th term} = 0 \) [From (1) and (2)]

Question. In an A.P., the first term is 25, nth term is − 17 and sum of first n terms is 60. Find ‘n’ and ‘d’, the common difference.
Answer: Here, the first term \( a = 25 \)
And the nth term = \( - 17 = l \)
Using \( T_n = a + (n - 1) d \), we have:
\( - 17 = 25 + (n - 1) d \)
\( \Rightarrow (n - 1) d = - 17 - 25 = - 42 \)
\( \Rightarrow (n - 1) d = - 42 \Rightarrow d = \left[ \frac{- 42}{n - 1} \right] \) ...(1)
Also, \( S_n = \frac{n}{2} [a + l] \)
\( \Rightarrow 60 = \frac{n}{2} [25 + (- 17)] \)
\( \Rightarrow 60 = \frac{n}{2} [8] \)
\( \Rightarrow 60 = 4n \Rightarrow n = \frac{60}{4} = 15 \)
From (1), we have
\( d = \frac{- 42}{15 - 1} = \frac{- 42}{14} = - 3 \)
Thus, \( n = 15 \) and \( d = - 3 \)

Question. In an A.P., the first term is 22, nth term is − 11 and sum of first n terms is 66. Find n and d, the common difference.
Answer: We have
1st term \( (T_1) = 22 \Rightarrow a = 22 \)
Last term \( (T_n) = - 11 \Rightarrow l = - 11 \)
Using, \( S_n = \frac{n}{2} [a + l] \), we have:
\( 66 = \frac{n}{2} [22 + (- 11)] \)
\( \Rightarrow 66 \times 2 = n [11] \)
\( \Rightarrow n = \frac{66 \times 2}{11} = 12 \)
Again using
\( T_n = a + (n - 1) d \)
We have:
\( T_{12} = 22 + (12 - 1) \times d \)
\( - 11 = 22 + 11d \) [\( \because \) nth term = − 11]
\( \Rightarrow 11d = - 22 - 11 = - 33 \)
\( \Rightarrow d = \frac{- 33}{11} = - 3 \)
Thus, \( n = 12 \) and \( d = - 3 \)

III. HOTS QUESTIONS

Question. Find the ‘6th’ term of the A.P. : \( \frac{2m + 1}{m}, \frac{2m - 1}{m}, \frac{2m - 3}{m}, .... \)
Answer: Here, \( a_1 = \frac{2m + 1}{m} \), \( a_2 = \frac{2m - 1}{m} \)
\( \therefore d = a_2 - a_1 \)
\( = \frac{2m - 1}{m} - \frac{2m + 1}{m} = \frac{2m - 1 - 2m - 1}{m} \)
\( = \frac{- 2}{m} \)
Now, \( a_n = a + (n - 1)d \)
\( \Rightarrow a_n = \left[ \frac{2m + 1}{m} \right] + (n - 1) \left[ \frac{- 2}{m} \right] \)
\( = \left[ \frac{2m + 1}{m} \right] + \left[ \frac{- 2n}{m} + \frac{2}{m} \right] - 1 \cdot \left[ \frac{- 2}{m} \right] \text{ [Wait, expanding via arithmetic]} \)
\( = \frac{2m + 1 - 2n + 2}{m} \)
\( = \frac{2m - 2n + 3}{m} \)
Thus, the \( n^{th} \) term is \( \left( \frac{2m - 2n + 3}{m} \right) \)
Again, we have \( a_n = \frac{2m - 2n + 3}{m} \)
\( \Rightarrow a_6 = \frac{2m - 2(6) + 3}{m} = \frac{2m - 12 + 3}{m} \)
\( = \frac{2m - 9}{m} \)
i.e., the 6th term is \( \left( \frac{2m - 9}{m} \right) \)

Question. If the numbers a, b, c, d and e form an A.P., then find the value of a – 4b + 6c – 4d + e
Answer: We have the first term of A.P. as ‘a’.
Let D be the common difference of the given A.P.,
Then :
\( b = a + D, c = a + 2D, d = a + 3D \) and \( e = a + 4D \)
\( \therefore a - 4b + 6c - 4d + e \)
\( = a - 4(a + D) + 6(a + 2D) - 4(a + 3D) + (a + 4D) \)
\( = a - 4a + 6a - 4a + a - 4D + 12D - 12D + 4D \)
\( = 8a - 8a + 16D - 16D = 0 \)
Thus, \( a - 4b + 6c - 4d + e = 0 \)

Question. If \( \frac{a^{n+1} + b^{n+1}}{a^n + b^n} \) is the arithmetic mean between ‘a’ and ‘b’, then, find the value of ‘n’.
Answer: Note : A.M., between ‘a’ and ‘b’ = \( \frac{1}{2} (a + b) \)
We know that :
A.M. between ‘a’ and ‘b’ = \( \frac{a + b}{2} \)
It is given that, \( \frac{a^{n+1} + b^{n+1}}{a^n + b^n} \) is the A.M. between ‘a’ and ‘b’
\( \therefore \frac{a^{n+1} + b^{n+1}}{a^n + b^n} = \frac{a + b}{2} \)
By cross multiplication, we get :
\( 2 [a^{n+1} + b^{n+1}] = [a^n + b^n] [a + b] \)
\( \Rightarrow 2a^{n+1} + 2b^{n+1} = a^{n+1} + ab^n + a^nb + b^{n+1} \)
\( \Rightarrow 2a^{n+1} - a^{n+1} + 2b^{n+1} - b^{n+1} = ab^n + a^nb \)
\( \Rightarrow a^{n+1} + b^{n+1} = ab^n + a^nb \)
\( \Rightarrow a^{n+1} - a^nb = ab^n - b^{n+1} \)
\( \Rightarrow a^n [a - b] = b^n [a - b] \)
\( \Rightarrow \frac{a^n}{b^n} = \frac{(a - b)}{(a - b)} = 1 \)
\( \Rightarrow \left( \frac{a}{b} \right)^n = \left( \frac{a}{b} \right)^0 \)
\( \Rightarrow n = 0 \)

Question. If pth term of an A.P. is \( \frac{1}{q} \) and qth term \( \frac{1}{p} \), prove that the sum of the first ‘pq’ terms is \( \frac{1}{2} [pq + 1] \).
Answer: Hint: \( p^{th} \text{ term} = \frac{1}{q} \Rightarrow a + (p - 1)d = \frac{1}{q} \) ...(1)
\( q^{th} \text{ term} = \frac{1}{p} \Rightarrow a + (q - 1)d = \frac{1}{p} \) ...(2)
Solving (1) and (2), \( d = \frac{1}{pq} \) and \( a = \frac{1}{pq} \)
Using \( S_n = \frac{n}{2} [2a + (n - 1)d] \), we get :
\( S_{pq} = \frac{pq}{2} \left[ 2 \left( \frac{1}{pq} \right) + (pq - 1) \times \frac{1}{pq} \right] \Rightarrow S_{pq} = \frac{1}{2} (pq + 1) \)

Question. If \( \frac{1}{b + c}, \frac{1}{c + a}, \frac{1}{a + b} \) are in A.P., prove that \( a^2, b^2, c^2 \) are also in A.P.
Answer: Hint:
\( \frac{1}{c + a} - \frac{1}{b + c} = \frac{1}{a + b} - \frac{1}{c + a} \)
[ using the fact, that in an A.P.
(2nd term – 1st term) = (3rd term – 2nd term)]

Question. Solve the equation : 1 + 4 + 7 + 10 + ... + x = 287
Answer: Since, 4 – 1 = 3, 7 – 4 = 3, 10 – 7 = 3
\( \Rightarrow 1, 4, 7, 10, ..., x \) form an A.P.
\( \therefore a = 1, d = 3 \) and \( a_n = x \)
\( \because a_n = a + (n - 1)d \)
\( \Rightarrow x = 1 + (n - 1) 3 \) or \( x = 3n - 2 \)
Also, \( S_n = \frac{n}{2} (a + l) \)
\( \Rightarrow 287 = \frac{n}{2} (1 + x) \)
\( \Rightarrow 2(287) = n[1 + (3n - 2)] \)
\( \Rightarrow 574 = n[3n - 1] \)
\( \Rightarrow 3n^2 - n - 574 = 0 \)
Solving the above quadratic equation, we get
\( n = \frac{-(-1) \pm \sqrt{1 + 4 \times 3 \times 574}}{6} = \frac{1 \pm \sqrt{6888}}{6} \text{ [Wait, check arithmetic } \sqrt{6889} = 83 \text{]} \)
or \( n = \frac{1 \pm 83}{6} \)
\( \Rightarrow n = 14 \) or \( n = \frac{-41}{3} \)
But, negative \( n \) is not desirable.
\( \therefore n = 14 \)
\( x = 3n - 2 \)
Now, \( x = 3(14) - 2 = 42 - 2 = 40 \)
Thus, \( x = 40 \)

Question. Find three numbers in A.P. whose sum is 21 and their product is 231.
Answer: Let the three numbers in A.P. are:
\( a - d, a, a + d \)
\( \therefore (a - d) + a + (a + d) = 21 \)
\( \Rightarrow a - d + a + a + d = 21 \)
or \( 3a = 21 \Rightarrow a = 7 \)
Also, \( (a - d) \times a \times (a + d) = 231 \)
\( \therefore (7 - d) \times 7 \times (7 + d) = 231 \)
\( \Rightarrow (7 - d) (7 + d) \times 7 = 231 \)
\( \Rightarrow 7^2 - d^2 = \frac{231}{7} = 33 \)
\( \Rightarrow 49 - d^2 = 33 \)
or \( d^2 = 49 - 33 = 16 \)
\( \Rightarrow d = \pm 4 \)
Now, when \( d = 4 \), then three numbers in AP are : \( (7 - 4), 7, (7 + 4) \) i.e. 3, 7, 11.
When \( d = - 4 \), then three numbers in AP are : \( [7 - (-4)], 7, [7 + (-4)] \)
or 11, 7, 3