The Arithmetic sequence(A.S) DefinitionA sequence in which the difference between any two successive terms is non-zero constant. i.e. Where d NoteIf Tn=nthterm (general term)is an algebraic expression of first degree in n,then(Tn)is an A.S.IN WHICH THE COEFFICIENT OF n=d EXAMpLESDetermine whether is an A.S or not. If it is an A.S ,Find its common difference and its first three terms.Solution = It is an A.S where d = 5 T1=5(1)+3=8 ,T2=5(2)+3=13,T3=5(3)+3=18Note : is first degree in n ==============================================================Determine whether is an A.S or notSolution = = . It isn’t an A.SNote : is an exponential function in n It isn’t an A.S ======================================================(3) Determine whether ( 2,5,8,11,………) is an A.S or not. Solution ==========================================================================================Determine whether ( 1,2,3,5, …….) is it an A.S or not?Solution ≠T4The general form of an A.SA.S (a , a + d , a+2d , ……. L - 2d. L - d , L)a= First termd= DifferenceLLast term====================================================== The general term A.S (3,5,7,……………) Find its general term Solution a=3 d= 2n+1 ============================================================================================A.S (6,3,0,…………..) Find Solution a=6 , =d ===========================================================================================A.S ( 12,22,32,………) Find the order of 202 .Solution a=12d= 202 -12=19 =n-1=============================================================================================Find the number of terms of the A.S ( 2,8,14,………,68 )SolutionThe number of terms = order of the last term a=2 d= 2 +68 – 2=11 =n-1 The number of terms =12If (29,x , ….., 3x , 95) is an A. S , then Find : (i) The value of x (ii) the number of its terms (iii) (IV) form the end Solution d = x – 29 = 95 – 3 x 95 = d = 31 – 29 = 2 95 = 29 + ( n – 1 ) (2) A.S ( 95,93,……,31,29) a = 95 , d = -2=============================================================================================To get the first negative term solve Find the order and the value of the first negative term in an A.S (42,38,34, …..)Solutiona = 42 , d = - 442 + ( n – 1 ) ( - 4 ) 42 – 4 n + 4 46 11.5 n = 12To get the first positive term solve A.S ( -76,-71,-66, …………) Find the first positive term Solution a = -76 , d = 5-76 + ( n – 1 ) ( 5) -76 + 5 n - 5 5 n n n = 17 =============================================================================================Find the number of negative terms in A.S ( -47, -42 , -37, …………) Solution a = -47 , d = 5 - 47 + ( n – 1 ) ( 5 ) - 47 –5 n - 5 5 n n n = 10 the number of negative terms = 10Prove that zero doesn’t belong to A.S (91, 85, 79,……..) Solution a = 91 , d = - 6 91 + ( n – 1 ) ( - 6 ) 91 – 6 n + 6 6 n n = =============================================================================================Let the number of the terms of A.S ( 8 , a , b , …., h , 68 ) be 16, so Find the value of each of a , b , and h . Solution 8 + 15 d = 68 d = 4 If from the A.S ( 13 , from the A.S (19, , ….) Then Find n .SolutionA.S ( 13, a = 13, d A.S ( 19 ,22,……)a = 19, d If the ratio between from an A.S is 3 : 5 then Prove that the ratio between is 2 :3 .Solution 2 a Find the A.S in which . Solution 13=5+4 d A.S ( 5 , 7, 9,………)Find the A.S in which Solution (2) (2) - (1) 4d = 12 Subs . in (1) a + 6 = 15 A.S ( 9 , 12 , 15 , ………) =============================================================================================Find the A.S in which Solution (1) 2a + 5d =24 (2)Subs in (2) by (1) 2(23-8d) +5d =24 46 - 16d+5d=24 22 = 11 d Subs in (1) a=23 – 8A.S(7,9,11,……………………..)Find the A.S in which the sum of its third and fourth terms equals 7 and exceeds by 4 Solution a +2d + a +3d =7 2a+5d=7 (1) (a+8d)-(a+4d)=4 4d = 4 d = 1 Subs in ( 1 ) 2a+5=7 2a=2 a = 1A.S (1,2,3,……………….)===============================================================================================Find the A.S in which the sum of its equals 4 and the product of its equals 7 show that there are two A.S. Solution a + d + a + 4d=4 2a+5d=4∴a= (1) (a+2d)(a+5d)=7 (2) Subs in (2) by (1) ( 4 – d ) (4 + 5d ) =28 16+16d - 5 5 ( 5d - 6 ) ( d – 2 ) = 0 d = subs in (1) a = a= A.S A.S An increasing A.S in which the sum of its second and its third terms equal 15 and Find the A.S. Solution 2a+3d=15 (1) (2)Subs in (1) by (2) 2(15 - 4d) +3d =15 15 = 5d d = 3Subs in (2) A.S (3,6,9,…………….) The order of the middle term = Find the A.S such that the number of its terms is 21 and its middle term equal 32 and the sum of its last three terms equals 177. SolutionOrder of the middle term = (1) (2) (2)-(1) Subs in (1) A.S (2,5,8, ……………)Three numbers makes an A.S, their sum is 33 and their product is 792, Find these numbers. SolutionLet the number be: Three numbers are 4,11,18 ==============================================================================================Four numbers makes an A.S, their sum equals 36 and the sum of their squares equals 344, Find these numbers. SolutionLet the number be:A.S 4 324+20 20 d =-1 d = 1 The four numbers are 6,8,10,12Find the A.S in which . Solution (1) 2a+2nd-2d=16 (2) (3) – (3) Subs in (1) by (2) Subs in (4)A.S (11, 10Let the first and the third terms of an A.S be equal respectively to the second and the fifth terms in another A.S, then Prove that the fifth term of the first A.S is equal to the eighth term of the second A.S.SolutionA. (1) (2)Subs in (2) by (1) )A.S 1 =)A.S 2==============================================================================================An A.S in which , if z+n=m+y, then Prove that k+e=L+x .Solution The Arithmetic MeanA.Sb is called A.M of a and c===============================================================================================The ratio between two numbers is 4:5 and their A.M is 18, Find the two numbers. SolutionLet the two numbers be A.M = 18 The two numbers are 16 and 20.The Arithmetic mean of two numbers is 23 and their product is 493. Find the two numbers. SolutionLet the two numbers be x and yA.M = 23x=46 - y(1) xy= 493(2)Subs in (2) by (1)(46 – y ) y = 493 46 y – The two numbers are 17 and 29.===============================================================================================A.S Find a , b , and c.Solution A.S Find the value of x. Solution 2 A.S Solution (2) From (1) and (2) ===============================================================================================IF a and L are two numbersTo insert nA . between a and LTn+2Insert 19 A.M. between - 19 and 81. Solution 81=-19 +20 d19 A. IF a and L are two numbersA.S (a, a+d, a+2d, a+3d,……….L – 3d, L – 2d, L – d , L) 1st mean 2nd mean 3rd mean (n-2) mean (n-1) mean nth mean L = If we insert n A.M. between 5 and 41 and if the ratio between the sum of the first two means and the sum of the last two means is 4 : 19, then Find the A.S.SolutionA.S 190 + 57d = 328 - 12d 69d=138 d = 2A.S ( 5 , 7 , 9 ,………, 39 , 41 )IF we insert n arithmetic means between 2 and 72 such that the ratio between the fourth mean and the mean whose order is (n-3) is 11 :26, then find n.SolutionA.S ( 2, 2+d , 2+2d, ……….., 72-2d, 72-d, 72)Fourth mean = ( n-3 ) mean = =52+104d=22+11(n-3)d52+104d=22+11nd-33d 30+137d+11nd (1) 2+(n+2-1)d=72+nd+d =72 nd=70-d (2)subs. in (1) by (2) 30+137d =11(70-d) 30+137 =770 - 11d 148d = 740 d = 5 subs. in (2)5n = 70 – 5n = 13Another Solution:A.S ( 2, 2+d , 2+2d, 2+3d, 2+4d, ………., 72 - 4d, 72 - 3d, 72 - 2d, 72 - d, 72) 4th mean (n-3) mean (n-2) mean (n-1) mean n th mean52 + 104d=792 - 44d 148d=740d = 572 + 72=2+(n+2-1)(5) 70 = (n+1)5=13===============================================================================================Let x , y and z be three positive number such that Prove that log x is the A.M of log y and log z. Solution Log x is the A.M of log y and log z.Let (x,y,z) be in an A.S , Prove that Solution 2y = x + zL.H.S = = = ===============================================================================================Let a , b and c be in an A.S, Prove that Solution2b = a + cL.H.S = = ===============================================================================================The sum of three consecutive numbers of an A.S is 24 ,if we subtract 2 from the middle term, then the multiplicative inverse of the resultant numbers form another A.S Find the three numbers. SolutionLet the three numbers be a – d, a a + d a – d+a+a+d =24 3a = 24 a = 8A.A. =============================================================================================== The measure of one of its angles is the A.M of the measures of the other two angles, let the difference between the measure of the smallest and the largest angle equal 80, Find the measure of each angle. SolutionThere are in an A.S Let the three angles be a - d , a , a + d (a + d)-(a – d ) = 80 2 d = 80 d = 48a – d + a +a +d = 1803a=180The three angles are .=============================================================================================== The sum of n terms of an A.S Where : n = number of terms sum of n terms a = starting term“ not always L = last term = ===============================================================================================Find the sum of the first 15 terms of the A.S ( 32,29,26,……….). Solutiona = 32 , d = -3 , n = 15 ===============================================================Find the sum of the terms of the A.S ( 76,71,66,….-24). Solutiona = 76 , d=-5 , n = ?Number of terms = order of the last term -24 = -24 = 76+ N = ===============================================================Prove that: SolutionL.H.S = ===============================================================================================Find the sum of the first 25 terms of the A.S ( Solution Find the sum of the natural number between 100 and 170 such that each of term is divisible by 3. Solution 1st number divisible by 3is 102 = aLast number divisible by 3is 168 = LNumber of numbers that is divisible by 3= ===============================================================================================(47) The sum of 20 terms of the A.S (starting from its tenth term. Solution Order of the last term =10+20-1=29Another Solution: S20=[2×39+9×4]=1540 Find the sum of first 30 terms of the sequence where: Solution =============================================================== Find the number of terms which must be taken from the A.S (1,3,5,…) starting from its first term such that their sum equals 400. Solution a = 1 , d = 3-1= 2 How many terms should be taken from A.S (40, 36, 32,…) starting from its first term, such that their sum becomes 208? Explain your answer. Solution a = 40 , d = - 4 n n 21 n = 13 , n = 8It means that: ================================================================Find the greatest sum of the terms of the A.S (33, 31, 29,…). SolutionThe greatest sum = sum of the positive termsThe order of the last positive term33+ 33 S17=[2×33+16×-2]=289Find the smallest number of terms must be taken from the A.S (24,21,18,…) starting from its first term, such that the sum becomes negative. Solution a = 24 , d = - 3 51 ===============================================================================================If we want to take 25 terms from the A.S (50,48,46,….) such that their sum equal 400, Find the starting term. Solutiond=-2 , value of the starting term The starting term Find the A.S whose common difference is and the sum of its first 21 terms equal zero. Solution A.S ( 40,36,32,….)===============================================================================================Find the A.S whose first is 12 and its last term equal and the sum of its terms equal . Solution is an A.S in which and , Find the sequence and the sum of its first 20 terms. Solution Subs in (2) by (1) 2 Subs in (2)A.S If the sum of the first three terms from an A.S is and the sum of its and equals , then Find the sum of the first 20 terms. Solution (1) (2)Subs in (2) by (1) Subs in (1)A.S A decreasing A.S in which the sum of it’s is 13 and their product equals 40, Find the sequence and the sum of its 12 terms. Solution (2)Subs in (2) by (1) 169 d= Subs in (1) A.S ( 11,9.5,8,…..) Find the A.S in which the sum of the first 20 terms equals 820 and the arithmetic mean of its equals 21. Solution (1) (2)(1) – (2) Subs in (2) ===============================================================================================Find the A.S in which its second term is 23 and its term before the last equals 97 and the sum of its terms equal 2400. Solution (1) add (2) subt. d=2 A.S(21,23,25,….) (60) Find the sum of the first n terms of the A.S (2,8,14,…) equals 4 times the sum of the first n terms of the A.S (20,21,22,…), so Find n. SolutionA.S(2,8,14,….) A.S(20,21,22,…) n = 79==========================================Let A.S (-12,-5,2,….,107):The number of its termsThe sum of this A.SThe sum of the terms which have even orders. Solution 107= 107=-12+ A.S ( -5,9,16,…., 107) n = 9 Find the A.S in which the sum of its first n terms is given by : Solution A.S(9,15,21,……….)============================================================Positive A.S in which and Find the sum of its first 10 terms. Solution (1) subs.in(1) A.S (5,20,35,…………………) Find the A.S for which the sum of its first seven terms equals 245 and the sum of its next seven terms equal 98. Solution (1) (2)Multiply (1) by (2) subt (3)d=-3subs in (1)A.S (44,41,38,…..)(64) How many terms are to be taken starting from the first term of the sequence such that the ratio between the sum of its first terms and the sum of the other terms is 5:34. SolutionT1=4(1)+3=7,T2=4(2)+3=11,T3=4(3)+3=15 A.S(7,11,15,……….)Let the number of the terms be 3 n:xxx5 34 T1 Tn T3n 65+26n=25+30n 40=4n n=10The number of terms = 3n = 3 (64)An A.S in which Find the A.S , then Find the sum of its first 3n terms. Solution (1) (2) (3)Multiply (2) by (2) (4) (4) - (3) (5) (1) - (5) 5d = -7.5d= -1.5subs in (5) A.S(34,32.5,31,……….)Subs in (2) 34 19.5=1.5nn = 13 ============================================================================================The number of terms of an A.S is 2n, let and the sum of its last n terms equals 600, so Find the A.S and the sum of the first n terms. Solutionxxxx First n term last n term n = 10 (1) (2)(2) – (1) 9 d = 4Subs in (1) a = 2A.S( 2,6,10,…….78) ===============================================================================An A.S whose terms are natural numbers the sum of its first ten terms where 240 , and its third term equals 16, Find this sequence. Solution A=16 - 2d (1)2402404848163.2 A.S(8,12,16,….)===============================================================================================A man saves at the end of a year L.E.7500 and he saves every year L.E.1500 more than the preceding year Find:How much dose he saves in the 17 th year.How much dose he saves in first 17 year. SolutionA.S ( 7500,9000,10500,………)A man was owed by L.E 48000, he decided to repay his debt on twenty installments yearly forming A.S after he paid five installments , he died and he was owed by of the debt, what was the first installment? Solution (1) (2) 15d = 960 d=64subs in (2) A tank holds 625 liter, a tap was fixed on it to pour water in the tank by rate 40 liters in the first hour and with an increase of 5 liters on every hour than the hour preceded it, after how many hours with the tank be filled? Solution A.S n = 10After 10 hours it will be filled.If the numbers of the neighboring hours in one side of a street are 1,2,3,……..,49, Find the number of the house which the sum of numbers of the houses preceded to it = the sum of number of the house followed it . Solution Let the number of the house be x 1,2,3,…….., x-1, x , x+1, ………., 48,49 x = 35The number of the house is 35. SUMMARY(1)A.S.(a,a+d,a+2d,a+3d,…….) where d If is an algebraic expression of first degree in n , Then in an A.S in which the coefficient of n = d. The general form of an A.SA.S (a , a + d , a+2d , ……. L - 2d. L - d , L)a= First termd= DifferenceLLast term======================================================(2)The general term ifTn=l n=+1(3)The number of terms = order of the last ter(4)To get the first "last"negative term Solve 0(5) To get the first"last"positive termSolve Tn˃o (6) The order of the middle term = (7)Arithmetic mean(A.M.)A.S b is called A.M of a and c i.e.2T2=T1+T3(8)IF a and L are two numbers to insert n A.Ms. between a and L L=Tn+2(9)IF a and L are two numbers A.S (a,a+d, a+2d, a+3d,………….,L-3d, L-2d, L-d, L) First second third n-2( n-1 )th nth Meanmean mean mean mean mean(10)The sum of n terms of an A.S Where : n = number of term Sum of n terms a = starting term“ not always L = last term = (11)THREE NUMBERS IN AN A.S. a- d, a ,a+d(12)FOUR NUMBERS IN AN A.S. a-3d , a-d ,a+d , a+3d(13)THE MAXIMUM SUM=THE SUM OF THE POSITIVE TERMS