Maths (O) 1995 PI Question 5 C

hacktavist

Kinda like the other thread help is apreciated! I know if i dont figure this out the same thing will come up!

In an arithmetic series, the tenth term, T10, is 19 and the sum to ten terms, S10 is 55.
Find the first term and the common difference.

Anyone? Only 2 days left!

Crumbs

S10 = 55
T10 = 19

Therefore, S9 = 36.

The sum to n terms, Sn = n/2[2a+(n-1)d]

Work that out for S9 and S10 and then solve the two equations to get a and d.

rasher_b2

you could use T10=19 so a+9d=19
and then work it out with the other one using simultaneous equations.

Sev

Right well an arithmetic sequence is a sequence of numbers such that there is a constant difference,d, between each.

So T2 = T1 + d, T3 = T2 + d, and so on

Therefore, T3 = T1 + 2d, likewise T9 = T1 + 8d

You get the idea.

So to get the sum of the first 10 numbers in an arithmetic sequence its simply

S10 = T1 + T2 + ..... + T10
= T1 + (T1 + d) + (T1 + 2d) + ..... + (T1 + 9d)

= 10(T1) + d (1 + 2 + 3 + .... + 9)

now there is a formula given on the maths course for the sum of an increasing series of integers = N(N+1)/2

so the sum of integers from 1 to 9 is 9(9+1)/2 = 45

so we get an expression for the sum of the first 10 numbers in the sequence to be

S10 = 10(T1) + 45d = 55

We also know that

T10 = T1 + 9d = 19

And now we just have two simultaneous equations, and by multiplying accross the latter by 10...

10T1 + 90d = 190
10T1 + 45d = 55

---

45d = 135
=> d = 3

And if d = 3, then T1 = -8

Camogie Playa

i remember trying to that sum too, but failed miserably!but im even more confused now

hacktavist

Ahh didnt think to use similtaneous equations! Chears lads.