## yasmeenabutineh 3 years ago What is the sum of the arithmetic sequence 3, 9, 15…, if there are 34 terms?

1. virtus

s= n/2(2a+(n-1)d) where s = sum of progression n = number of terms a = value of first term d = difference between 2 terms so for 3, 9,15 n= 34 since there are 34 terms a= 3 d= 6 so now lets input these into our formula s=(34/2) x {2(3) +[(34-1)x6]} = 17 x 210 =3570

2. Martin_An

I don't know if this is a typo but in the above formula "a" should not be multiplied by two. The correct answer is 3468. Here's why. The initial term is n_0=3. Each successive term grows by 6 and there are 33 more additional terms in the sequence. Therefore the last term is actually 3 + 33(6)=201. The sum of the first and last term is therefore 204. When you consider the other pairings (e.g. n_1 + n_32 and so on) their sums don't change since the first term increases by 6 while the second decreases by 6. And since we have 34 members in the sequence, there are 17 such pairs each with a sum of 204. Therefore the sum of the whole sequence is 17x204=3468.