Succession 2o1a5r

ACTIVITIES FOR STRENGTHENING AND EXTENSION MATH. 3rd E.S.O. UNIT 3: SUCCESSION. Progression. 1.) -6,1,8,15 Given the succession ... a) Is an arithmetic progression? c) Calculate the general term. 1 - (-6) = 7 8-1 = 7 15 - 8 = 7 It is a p.a. a1 = -6 and d = 7 an = a1 + (n - 1) ⋅ d n = -6 + (n - 1) ⋅ 7 ⇒ n = 7n - 13 b) Write its first eight . -6,1,8,15,22,29,36,43 d) Calculate the term 11. How much is a20? a11 = 7 ⋅ 11-13 = 64-20 = 7 ⋅ 20-13 = 127 e) Is 197 a term of the sequence? What about 224? We proceed from the general term n = 7n - 13 solve the equation 197 = 7n - 13 ⇒ n = 30. Then 30 = 197 237 We see that n (position of the term in the sequence 7) is not a natural numb er. 224 is not a term of this progression. Similarly, we solve the equation 224 = 7n - 13 ⇒ n = 2.) Given the sequence -5 -3, -2, -1, ... February 2 a) Is an arithmetic progression? 5 1 -2 2 2

-

=

2

2 to 3 January May 1 is a pa a1 = - and d = - (-2) = 2 2

3 1 -1 = 2 2 b) Calculate its general term. c) Calculate the a12. How much is a60? an = a1 + (n - 1) ⋅ d n = - 5 1 5n1 n-6 + (n - 1) ⋅ = - + - ⇒ n = 2 February 222 2 a12 = 12-6 = 3 2 60 = 60-6 = February 27 3.) In an arithmetic progression know the words to 8 = 29 and a11 = 44. Calculat e: a) The difference of the sequence and the first term. n - k = (n - k) ⋅ d a11 - to 8 = 3d ⇒ 1944-1929 ⇒ d = 3d = 5-8 - a1 = 7 ⇒ 29 - a 1 = 7 ⋅ 5 ⇒ a1 = 29-35 ⇒ a1 = -6 b) The general term succession.

an = a1 + (n - 1) ⋅ d n = -6 + (n - 1) ⋅ 5 ⇒ an = 5n - 11 mate-alceda.blogspot.com Page 1 of 3 4.) Knowing the to 4 = 1 and a10 = 10, an arithmetic progression, is the g eneral term. Solution: n = 5.) Calculates the sum of the first 80 multiples of 4. 3n - 10 2 Consecutive multiples of 4 is a p.a. general term an = 4n Sn = S80 S80 ⋅ No 2 (a1 + A80) a1 = 4-80 = 320 = 80 ⋅ 2 (4 + 320) = 2 ⋅ 80 = 12 960 (A1 + an) 6.) How many multiples of 3 between the first 172 natural numbers? Solution: Multiple 57 8.) Given the sequence 16, -8.4, -2, ... a) Is an arithmetic progression? -8 To 16 = -24 4 - (-8) = 12 Not a p.a. because the difference between two consecutive is not constant. b) Is a geometric progression? -8: April 16 = -0.5 (-8) = -0.5 is a pg a1 = 16 and r = - -2: 4 = -0.5 c) Write the first eight . February 1 16, -8.4, -2.1, -1 1 -1, 248 e) Calculate the ninth term. How much is a12? 9 -1 d) Calculate the general term. an = a1 ⋅ r n - 1 -1

n = 16 ⋅

2

n-1 -1 to 9 = 16 ⋅ January 16 -1 8

2 ⋅

= 16 = = 256 16

2

-1

a12 = 16 ⋅

2

12-1 = 1128 9.) Given the sequence 3 24 8, 1,,,, ... 2 March 9, 1927 b) Calculate the right. Calculates its general term. a) Is a geometric progression? 1: 32 = 23 2 2 1 = 3 2 3 3 is a P.G. a1 = and r = 42 2 2 3: = 93 3842: = 27 9 3 an = a1 ⋅ rn - 1 n = 3 n-1 2

=

2 2

⋅

3

3

n-2 ACTIVITIES FOR STRENGTHENING AND EXTENSION MATHEMATICS. 3rd E.S.O. UNIT 3: SUCCESSION. Progression. 10.) In a geometric progression the third term is a3 = 52 and the reason is r = 2. Calculate: a) The first term. b) The general term. n = rn-k to 52 k a3 = r 3 -1 r 2 = ⇒ 3 ⇒ ⇒ a1 = 22 = 13 a1 a1 a1 c) The sum of the first ten . an = a1 ⋅ r n - 1 to n = 13 ⋅ 2n - 1 Calculate the first ten 13,26,52,104,208, 416,832,1664,3328,6656, Their su m is 13 299 11.) In a fifth term geometric progression is 2 and a2 = 54. Calculate the ratio and the general term. n = r n-k ak a5-2 5-2 1 1 = r ⇒ 5 = r 3 = r3 ⇒ ⇒ ⇒ r = r3 = a2 a2 54 27 3 To calculate the ge neral term we need to compute a1 = a1 1 How to 2: r ⇒ a1 = 54: = 162 3 ⋅ n = a1 rn - 1 1

n = 162 ⋅

3

n-1 12.) In a geometric progression know the words to 5 = 112 and 8 = 896. Calculate a3. Solution: 3 = 28 13.) How many multiples of 3 between 120 and 512 numbers? Between 120 and 512 the first multiple of 3 is 120 and the last is 510. The mult

iples of 3 is a p.a. 3n general term and difference 3. To calculate the number o f Finally tér mino - first 510-120 +1 = + 1 = 131 different 3 Why add 1 to the formula? Solution: Multiple 131 14.) Find the first term of an arithmetic progression is known that the term is ranked 11 is double that ranks 7, and the difference in the progression is 0.5. a11 = 2 ⋅ 7 (data of the problem) On the other hand 7 = a1 + 6d

⇒ 7 = 4d a11 - to 7 = 4d

⇒ a1 =-2d = -1 to 7 = 4d

15.) A cyclist intends to cycle through all the towns in his province. The first day for 80 km, and each day remaining runs 5 km more than the previous. Knowing that planned 665 km, how man y days will run its course? Solution: 7 days mate-alceda.blogspot.com Page 3 of 3