Arithmetic Sequence Formulas:
A sequence is said to be an arithmetic sequence if the difference between the consecutive terms is always constant. This constant is called the common difference denoted by ‘d’ and is given as-
| d = an – an-1 |
Arithmetic Sequence is also called Arithmetic Progression.
General Form of Arithmetic Sequence:
If ‘a’ be the first term and ‘d’ be the common difference of the arithmetic sequence then the general form of the arithmetic sequence is a, a + d, a + 2d, a + 3d, and so on.
nth– term of Arithmetic Sequence:
If ‘a’ be the first term and ‘d’ be the common difference of an arithmetic sequence, then its nth term is given by-
| an = a + (n – 1) d |
mth term from the end:
If there are n terms in a sequence, then the mth term from the end is equal to (n – m + 1)th from the beginning.
Selection of terms in Arithmetic Sequence:
If we have to select a certain number of terms in an arithmetic sequence, it is better to select the terms in the following ways-
| Number of Terms | Terms to be Selected | Common Difference | Sum of the Terms |
|---|---|---|---|
| 3 | a – d, a, a + d | d | 3a |
| 4 | a – 3d, a – d, a + d, a + 3d | 2d | 4a |
| 5 | a – 2d, a – d, a, a + d, a + 2d | d | 5a |
| 6 | a – 5d, a- 3d, a – d, a + d, a + 3d, a + 5d | 2d | 6a |
Sum n-terms of Arithmetic Sequence:
Let Sn be the sum of n terms of an arithmetic sequence, then-
| Sn = n/2 [2a + (n – 1) d] ……….(i) |
If a be the first term and an be the last term, then-
| Sn = n/2 [a + an] ……….(ii) |
| Example- Which term of the sequence -3, 5, 13, ……….. is 125. Solution- The given sequence is, -3, 5, 13, ……… Here, a = -3, d = 5 – (-3) = 8 Let an = 125 ⇒ a + (n – 1) d = 125 ⇒ -3 + (n – 1) 8 = 125 ⇒ (n – 1) 8 = 128 ⇒ (n – 1) = 16 ⇒ n = 17 ∴ 125 is the 17th term of the given arithmetic sequence. |
| Example- Is 151 a term of the series 7, 10, 13, …….. Solution- The given sequence is, 7, 10, 13, ……….. Here, a = 7 d = 10 – 7 = 3 Let an = 151 ⇒ a + (n – 1) d = 151 ⇒ 7 + (n – 1) 3 = 151 ⇒ (n – 1) 3 = 144 ⇒ n – 1 = 48 ⇒ n = 49 ∴ 151 is the 49th term of the arithmetic sequence. |
| Example- The 6th and 15th terms of an arithmetic sequence are 23 and 59 respectively. Find its 50th term. Solution- Let ‘a’ and ‘d’ be the first term and common difference of arithmetic sequence respectively. Now, a6 = 23 ⇒ a + 5d = 23 ……….(i) and a15 = 59 ⇒ a + 14d = 59 ……….(ii) Subtract (i) from (ii):- 9d = 36 ⇒ d = 4 Substitute d = 4 in equation (i):- a + 5(4) = 23 ⇒ a = 3 Now, a50 = a + 49d ⇒ a50 = 3 + 49(4) ⇒ a50 = 3 + 196 = 199 |
| Example- If p times the pth term of an arithmetic sequence is equal to q times the qth term, then prove that its (p + q)th term is zero. Solution- Let ‘a’ be the first term and ‘d’ be the common difference of an arithmetic sequence. Now, p ap = q aq ⇒ p [a + (p – 1) d] = q [a + (q – 1) d] ⇒ pa + p (p – 1) d = qa + q (q – 1) d ⇒ pa – qa = q (q – 1) d – p (p – 1) d ⇒ (p – q) a = (q2 – q – p2 + p) d ⇒ (p – q) a = [(q2 – p2) + (- q + p)] d ⇒ (p – q) a = [(q – p) (q + p) + (p – q)] d ⇒ (p – q) a = [- (p – q) (p + q) + (p – q)] d ⇒ (p – q) a = (p – q) [- (p + q) + 1] d ⇒ a = [- (p + q) + 1] d ⇒ a = – [p + q – 1] d ⇒ -a = (p + q – 1) d ………(i) Now, ap+q = a + (p + q – 1) d ⇒ ap+q = a + (-a) [using (i)] ⇒ ap+q = 0 |
| Example- If the nth term of an arithmetic sequence is n and its nth term is m. Then show that its (m + n)th term is zero. Solution- Let ‘a’ be the first term and ‘d’ be the common difference of an arithmetic sequence. Now, am = n ⇒ a + (m – 1) d = n ……….(i) and an = m ⇒ a + (n – 1) d = m ……….(i) Subtract (i) from (ii):- (n – m) d = m – n ⇒ (n – m) d = – (n – m) ⇒ d = -1 Substitute d = -1 in equation (i):- a + (m – 1) (-1) = n ⇒ a = n + (m – 1) Now, am+n = a + (m + n – 1) d ⇒ am+n = (n + m – 1) + (m + n – 1) (-1) ⇒ am+n = 0 |
| Example- Three numbers are in an arithmetic sequence. The sum of the numbers is 27 and their product is 585. Find the numbers. Solution- Let three numbers in an arithmetic sequence be a – d, a, a + d. Now, a – d + a + a + d = 27 ⇒ 3a = 27 ⇒ a = 9 Also, (a – d) (a) (a + d) = 585 ⇒ (a2 – d2) a = 585 ⇒ (92 – d2) 9 = 585 ⇒ 81 – d2 = 65 ⇒ d2 = 16 ⇒ d = 4 ∴ three numbers in an arithmetic sequence are 5, 9, 13 or 13, 9, 5. |
| Example- Find four numbers in an arithmetic sequence, whose number is 24 and the greatest of which is three times the least. Solution- Let four numbers in an arithmetic sequence be a – 3d, a – d, a + d, a + 3d. Now, a – 3d + a – d + a + d + a + 3d = 24 ⇒ 4a = 24 ⇒ a = 6 Now, a + 3d = 3 (a – 3d) ⇒ a + 3d = 3a – 9d ⇒ 2a = 12d ⇒ a = 6d ⇒ d = 1 ∴ four numbers in an arithmetic sequence are 3, 5, 7, 9. |
| Example- The sum of three terms of an arithmetic sequence is 30. The product of the first and the third terms exceeds 9 times the second term by unity. Find the numbers. Solution- Let three numbers in an arithmetic sequence be a – d, a, a + d. Now, a – d + a + a + d = 30 ⇒ 3a = 30 ⇒ a = 10 Now, (a – d) (a + d) – 9a = 1 ⇒ a2 – d2 – 9a = 1 ⇒ 102 – d2 – 9 (10) = 1 ⇒ 10 – d2 = 1 ⇒ – d2 = – 9 ⇒ d = ± 3 ∴ three numbers in an arithmetic sequence are 7, 10, 13. |
| Example- The angles of a quadrilateral are in an arithmetic sequence and the common difference is 30°. Find the measures of the angles. Solution- Let four angles of a quadrilateral in an arithmetic sequence be a, a + d, a + 2d, a + 3d. With d = 30° Now, a + a + d + a + 2d + a + 3d = 360° ⇒ 4a + 6d = 360° ⇒ 4a + 6 (30°) = 360° ⇒ 4a = 360° – 180° ⇒ a = 45° ∴ angles in arithmetic sequence are 45°, 75°, 105°, 135°. |
| Example- The sum of three numbers in an arithmetic sequence is 15 and the sum of their squares is 93. Find the numbers. Solution- Let three numbers in an arithmetic sequence be a – d, a, a + d. Now, a – d + a + a + d = 15 ⇒ 3a = 15 ⇒ a = 5 Now, (a – d)2 + a2 + (a + d)2 = 93 ⇒ 2 (a2 + d2) + a2 = 93 ⇒ 2 (52 + d2) + 52 = 93 ⇒ 50 + 2d2 + 25 = 93 ⇒ 2d2 = 18 ⇒ d2 = 9 ⇒ d = ± 3 ∴ three numbers in an arithmetic sequence are 2, 5, 8 or 8, 5, 2. |
| Example- Find the sum of the following series- (i) 13 + 21 + 29 + ……….. + 181. (ii) 4 + 7 + 10 + ……….. to 40 terms. Solution- (i) The given series is, 13 + 21 + 29 + ……….. + 181 Here a = 13 d = 21 -13 = 8 and an = 181 a + (n – 1) d = 181 ⇒ 13 + (n – 1) 8 = 181 ⇒ (n – 1) 8 = 168 ⇒ (n – 1) = 21 ⇒ n = 22 ∴ Sn = n/2 (a1 + an) ⇒ S22 = 22/2 (13 + 181) ⇒ S22 = 11 (194) = 2134 (ii) The given series is, 4 + 7 + 10 + ……….. to 40 terms Here a = 4 d = 7 – 4 = 3 and n = 40 Now, Sn = n/2 [2a + (n – 1) d] ⇒ S40 = 40/2 [2 (4) + (40 – 1) 3] ⇒ S40 = 20 (8 + 117) ⇒ S40 = 20 (125) ⇒ S40 = 2500 |
| Example- Find the sum to p terms of the sequence whose pth term is 2 – 3p. Solution- ap = 2 – 3p Here, a1 = 2 – 3 (1) = -1 ∴ Sp = p/2 (a1 + ap) ⇒ Sp = p/2 (- 1 + 2 -3p) ⇒ Sp = p/2 (1 – 3p) |
| Example- Find the sum of all multiples of 9 between 100 and 900. Solution- Multiples of 9 between 100 and 900 are, 108, 117, 126, ………., 900 It is an arithmetic progression with a = 108, d = 9, an = 900 a + (n – 1) d = 900 ⇒ 108 + (n – 1) 9 = 900 ⇒ (n – 1) 9 = 792 ⇒ (n – 1) = 88 ⇒ n = 89 ∴ Sn = n/2 (a1 + an) ⇒ S89 = 89/2 (108 + 900) ⇒ S89 = 89/2 (1008) ⇒ S89 = 89 (504) ⇒ S89 = 44856 |
| Example- Find the sum of the first 20 terms in A.P. in which the third term is 7 and the seventh term is two more than thrice its third term. Solution- Let a be the first term and d be a common difference. Now, a3 = 7 ⇒ a + 2d = 7 ……….(i) and a7 = 2 + 3a3 ⇒ a + 6d = 2 + 3 (7) ⇒ a + 6d = 23 ……….(ii) Subtract (i) from (ii):- 4d = 16 ⇒ d = 4 Substitute d = 4 in equation (i):- a + 2 (4) = 7 ⇒ a = -1 ∴ Sn = n/2 [2a + (n – 1) d] ⇒ S20 = 20/2 [2 (-1) + (20 – 1) (4)] ⇒ S20 = 10 (-2 + 76) ⇒ S20 = 740 |
| Example- The sum of n terms of two arithmetic series are in the ratio (7n + 1) : (4n + 27). Find the ratio of their mth terms. Solution- Sn/S’n = (7n + 1)/(4n + 27) ⇒ (n/2) [2a1 + (n – 1) d1]/(n/2) [2a2 + (n – 1) d2] = (7n + 1)/(4n + 27) Replace n by 2m – 1 ⇒ [2a1 + (2m – 1 – 1) d1]/[2a2 + (2m – 1 – 1) d2] = [7 (2m – 1) + 1]/[4 (2m – 1)+ 27] ⇒ [2a1 + (2m – 2) d1]/[2a2 + (2m – 2) d2] = (14m – 6)/(8m + 23) ⇒ [a1 + (m – 1) d1]/[a2 + (m – 1) d2] = (14m – 6)/(8m + 23) ⇒ am/a’m = (14m – 6)/(8m + 23) |
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