A brief note on Sequences, Series and Arithmetic Progressions . Read it before a test. Try each question at the end before you open its answer.
Sequences and series A sequence is a function on 1 , 2 , 3 , … 1, 2, 3, \dots 1 , 2 , 3 , … ; its n n n th term is a n a_n a n . A formula gives a n a_n a n at once. A recurrence builds each term from earlier ones. A series adds the terms: S n = a 1 + ⋯ + a n = ∑ k = 1 n a k S_n = a_1 + \dots + a_n = \sum_{k=1}^{n} a_k S n = a 1 + ⋯ + a n = ∑ k = 1 n a k . a 1 = S 1 a_1 = S_1 a 1 = S 1 , and a n = S n − S n − 1 a_n = S_n - S_{n-1} a n = S n − S n − 1 for n ≥ 2 n \geq 2 n ≥ 2 . Arithmetic progressions Each term is the one before plus d d d : a n = a + ( n − 1 ) d a_n = a + (n - 1)d a n = a + ( n − 1 ) d . From the end: the n n n th term from the end is l − ( n − 1 ) d l - (n - 1)d l − ( n − 1 ) d . S n = n 2 ( 2 a + ( n − 1 ) d ) = n 2 ( a + l ) S_n = \dfrac{n}{2}\left(2a + (n - 1)d\right) = \dfrac{n}{2}(a + l) S n = 2 n ( 2 a + ( n − 1 ) d ) = 2 n ( a + l ) . A sequence is an AP exactly when a n a_n a n is linear in n n n , or S n S_n S n is a quadratic in n n n with no constant term. Adding, subtracting, multiplying or dividing every term by a constant keeps an AP. Choose three numbers as a − d , a , a + d a - d, a, a + d a − d , a , a + d , and four as a − 3 d , a − d , a + d , a + 3 d a - 3d, a - d, a + d, a + 3d a − 3 d , a − d , a + d , a + 3 d . Arithmetic means The AM of a a a and b b b is a + b 2 \dfrac{a + b}{2} 2 a + b ; a a a , b b b , c c c are in an AP exactly when 2 b = a + c 2b = a + c 2 b = a + c . m m m means between a a a and b b b : d = b − a m + 1 d = \dfrac{b - a}{m + 1} d = m + 1 b − a , and the j j j th mean is a + j d a + jd a + j d . The m m m means add up to m × a + b 2 m \times \dfrac{a + b}{2} m × 2 a + b . A worked example An AP has a 2 = 11 a_2 = 11 a 2 = 11 , and the sum of its first 9 9 9 terms is 207 207 207 . Find its first term and common difference.
Use the sum. S 9 = 9 2 ( a 1 + a 9 ) S_9 = \dfrac{9}{2}(a_1 + a_9) S 9 = 2 9 ( a 1 + a 9 ) , and a 1 + a 9 = 2 a 5 a_1 + a_9 = 2a_5 a 1 + a 9 = 2 a 5 , because they are the same distance from the middle. So S 9 = 9 a 5 S_9 = 9a_5 S 9 = 9 a 5 . The middle term. 9 a 5 = 207 9a_5 = 207 9 a 5 = 207 , so a 5 = 23 a_5 = 23 a 5 = 23 . The difference. a 5 − a 2 = 3 d = 12 a_5 - a_2 = 3d = 12 a 5 − a 2 = 3 d = 12 , so d = 4 d = 4 d = 4 . The first term. a = a 2 − d = 7 a = a_2 - d = 7 a = a 2 − d = 7 . Practice with new numbers
A program to try The program checks the formulas of this note on one AP: the n n n th term, the sum, recovering the terms from the sums, and the middle term as a mean.
Run it. Do the formulas agree with the listed terms?
1. Change d to 5. Predict the sum of the 15 15 15 terms first.
Mistakes to avoid a + n d a + nd a + n d for the n n n th term. It is a + ( n − 1 ) d a + (n - 1)d a + ( n − 1 ) d . a n = S n − S n − 1 a_n = S_n - S_{n-1} a n = S n − S n − 1 at n = 1 n = 1 n = 1 . Use a 1 = S 1 a_1 = S_1 a 1 = S 1 . Dividing by m m m for m m m means. There are m + 1 m + 1 m + 1 steps. Rejecting a second root for n n n . Two values can both be right when some terms add up to zero. b 2 = a c b^2 = ac b 2 = a c for an AP. That is the geometric test. An AP needs 2 b = a + c 2b = a + c 2 b = a + c . Check yourself Find the 15 15 15 th term of the AP − 4 , 1 , 6 , … -4, 1, 6, \dots − 4 , 1 , 6 , … . Stuck? Show a hint Show the answerThe partial sums of a sequence are S n = 5 n 2 S_n = 5n^2 S n = 5 n 2 . Find a 3 a_3 a 3 . Stuck? Show a hint Show the answerFind the sum 9 + 13 + 17 + … 9 + 13 + 17 + \dots 9 + 13 + 17 + … to 12 12 12 terms. Stuck? Show a hint Show the answerInsert 2 2 2 arithmetic means between − 5 -5 − 5 and 13 13 13 . Stuck? Show a hint Show the answerFour numbers in an AP add up to 40 40 40 . What do the two middle ones add up to? Stuck? Show a hint Show the answer