Now to find the sum of the terms in an Arithmetic sequence. The set of a, a + d, a + 2d, a + 3d…….., a + (n - 1)d is an Arithmetic sequence but,….. The sum of a + (a + d) + (a + 2d) +…….+ a + (n - 1)d is an Arithmetic series A calculation to find the sum of the terms in series is needed! First of all, let un = a + (n - 1)d = l This is needed to create the calculation.

 2Sn = n(a + l) So, Sn = ½n(a + l)  Sn = ½n[a + a + (n - 1)d] i.e. This is just written in reverse! 1 and so on 2nd last term last term Sn = a + (a + d) + (a + 2d) +...+ (l - 2d) + (l - d) + l 1 Now, Sn = l + (l - d) + (l - 2d) +…+ (a + 2d) + (a + d) + a Also, 2 So, 2Sn = (a + l) + (a + l) + (a + l) +…+ (a + l) + (a + l) + ... There are n of these 1 2 +  2Sn = n(a + l) For this sum you need to know the first & last terms of the series So, Sn = ½n(a + l) since l = a + (n - 1)d And,  Sn = ½n[a + a + (n - 1)d] For this sum you need to know the first term & the common difference i.e. Sn = ½n[2a + (n - 1)d]

The whole expression has been multiplied by This is how to develop the formula to find the sum of the terms in a Geometric sequence: If the set of terms a, ar, ar2, ar3, ar4, …, arn-1 …. is a Geometric sequence, then The sum of a + ar + ar2 + ar3 + ar4 + …+ arn-1 +….. is a Geometric series. So, Sn = a + ar + ar2 + ar3 + ar4 + …+ arn-1 +…. The next step is tricky! rSn = ar + ar2 + ar3 + ar4 + …+ arn-1 + …+ arn + …+ The whole expression has been multiplied by Sn - rSn = a - arn Because all the terms cancel each other out! Sn (1 - r) = a(1 - rn) r By factorising! On BOTH sides! a(1 - rn) So, Sn = Move (1 - r) to the right by dividing. a(rn - 1) Sn = (r - 1) (1 - r) OR This one is used when r < 1 This one is used when r > 1

We can use this one because the 12th term will be the last one. Now for some examples……. First, some Arithmetic ones! 1. The first term of an Arithmetic series is 5 and the 12th term is -14. Find the sum of the first twelve terms. Sn = ½n(a + l) You have to remember this!  We can use this one because the 12th term will be the last one. S12 = ½  12 (5 - 14) n = 12 l = -14 S12 = 6  -9 a = 5 S12 = -54 2. Find the sum to 15 terms of 3 + 7 + 11 + 15 + ….. This is an Arithmetic series! Sn = ½n[2a + (n - 1)d] n = 15 d = 4 a = 3 S15 = ½  15[2  3 + (15 - 1)4] S15 = ½  15  62 S15 = 465

Same answer but a lot easier the other way! 3. Find the sum to 7 terms of the geometric series 2 + 6 + 18 + ……... a(1 - rn) Sn = (1 - r) a(rn - 1) Sn = (r - 1) OR It’s either It’s this one, because…. r > 1 a = 2 r = 3 n = 7 So, If you used S7 = 2186 Same answer but a lot easier the other way! That was easy. Now for a tough one!

  + + 5n + 5n Easy! & Now for the hard one! 4. The sequence 8, 14, 32, ……. has its nth tern 3n + 5. Find (i) the fourth term and (ii) the sum of the first n terms (i) the fourth term We are told that Un = 3n + 5 So U4 = 34 + 5  Easy! Now for the hard one! U4 = 86 (ii) the sum of the first n terms This is a tricky one! Because the series is neither Arithmetic NOR Geometric Now U1 = 31 + 5 & U2 = 32 + 5 Now add vertically 32 - 14  14 - 8 & U3 = 33 + 5 U4 = 34 + 5 It needs a SPECIAL method to solve! And so on  Sn = Sum of a Geometric series Sum of n terms each of which is 5 + + 5n So, Sn = where + 5n a = 3 So, Exercise 9G has plenty more! r = 3