Arithmetic progressions.

1. The sum, Sn, ofthe first n terms of a series is given by S; = 2n(n - 2).

(a) Find, in terms ofn, the nth term of the series. ( an = 4n-6)

(b) Hence, show that the series is an A.P.

(c) For the AP in (b) above, identify (i) the first term (ii) the common difference.( a = -2, d = 4) [CAPE 2005]

fh .1+1+1+ .2. (a) Verify thatthe sum, Sn,o t e senes"2 23 25 ..., to n terms, IS

S; = t(1- 2;n)'

(b) Three consecutive terms (x - d), x, and (x + d), d > 0, of an arithmetic series have sum 21and product 315. Find the value of

(i) x and (ii) the common difference (x = 7; d = 2) [CAPE 2006]

3. An AP with 10 terms has rust term 60 and last term -120. Find the sum of ALL the terms.[CAPE 2009, paper03/B]

m

4. (i) Show that the terms of L In 3r are in arithmetic progression. (a = In 3; d = In 3)r=1

(ii) Find the sum ofthe first 20 terms of this series. (210 In 3)

2m

(iii) Hence, show that L In 3 r = (2m2 + m) In 3.r=1

[CAPE 2006]

5. The first four terms of an AP are 2, 5, (2x +y + 7) and (2x - 3y) respectively, where x and yare constants. Find the values ofx and y (x = t;y = -5.) [CAPE 2004]

6. An AP has first term 3 and common difference d.

The sum of the rust six terms is 48 and the sum ofthe rust n terms is 168.

Find (a) the value of d and (b) the value ofn. ( d = 2; n = 12) [CAPE 2003]

7. The sum, Sn, of the rust n terms of a series is given by S; = n(3n - 4). Show that the termsof the series are in arithmetic progression with common difference 6. [CAPE 2002]

8. (a) An arithmetic sequence has first term 100 and common difference 2.

(i) Write down the second and third terms. (102; 104 )

(ii) Given that the last term is 200, find the number of terms. (51)

(b) A tape dispenser has a length of tape wrapped round a circular cylinder. The length oftape in the first layer (nearest to the cylinder) is 100 mm. Each further layer is 2 mmlonger than the layer before. The outer layer has 200 mm of tape. Calculate the totallength of tape. (7650 mm)

9. (a) Find the sum of the 100 terms of the arithmetic series

3 + 7 + 11 + ...+ 399 (20 100)

(b) An arithmetic sequence U], U2, U3, •.• has rth tern u., where

u; = 8r-2

(i) Write down the value of u., U2, U3 and U4. (6, 14,22 and 30)

(ii) Using your answer to part(a), or otherwise, find the sum of the first 100 terms ofthissequence. (40200)

10. Consider the arithmetic series101 + 104 + 107 + 110 + ...+ 800.

(a) Show that there are 234 terms in this series.

(b) Find the sum of the arithmetic series. (105417)

(c) Find the sum of the even numbers in the series. (52884)

11. The nth term of an arithmetic sequence is u; where

us= 90 -3n

(a) Fnd the value of u, and the value of U2. [87,84] (2 marks)

(b) Write down the common difference of the arithmetic sequence. [- 3] (1 mark)

k

(c) Given that L u; = 0 fmd the value ofk. [k = 59]n=1

(3 marks)

12. The arithmetic series51 + 58 + 65 + 72 + ...+ 1444

has 200 terms.

(a) Write down the common difference of the series. [d = 7] (1 mark)

(b) Find the 101 st term of the series. [751] (2 marks)

(c) Find the sum of the last 100 terms of the series. [109750] (2 marks)

13. John's father gave him a loan of$10 800 to buy a car. The loan was to be repaid by 12unequal monthly instalments, starting with an initial payment of $P in the first month. Thereis no interest charged on the loan, but the instalments increase by $60 per month.

(i) Show that P = 570 (3 marks)

(ii) Find, in terms of n, 1 ~ n ~ 12, an expression for the remaining debt on the loan after Johnhas paid the n th instalment. [CAPE 2007] (3 marks)

14. The first term of an arithmetic sequence is 2 and the 20th term is 97. Obtain the sum of thefirst 50 terms. [6225] (4 marks)

15. In an arithmetic sequence the second term is 7 and the sum of the first five terms is 50.Find the common difference of this arithmetic sequence. [d = 3 ] (3 marks)

16. The lights in a street will remain lit for 16 hours during the night of 4th of January 2010.Every subsequent night the time for which they will remain lit will be decreased by 200seconds.

(i) By using a suitable arithmetic progression show that they will remain lit for 52 000seconds during the night of 1st February 2010. (6 marks)

(ii) Find the total number of hours for which the Government has to pay for street lightsduring the month of February 2010 (i.e.night 1st February up to and including the nightof 28th February). [1 380400 seconds = 383 hours] (3 marks)

17. A progression has a first term of 12 and a fifth term of 18.

(i) Find the sum of the first 25 terms if the progression is arithmetic. [750] (3 marks)

(ii) Find the 13th term if the progression is geometric. [40.5] (4 marks)