Thread: Arithmetic Progression help needed!

Hey, sorry to keep asking for help but:

1. The 20th term of an arithmetic progression is 10 and the 50th term is 70.
(i) Find the first term and the common difference.

(ii) Show that the su of the first 29 terms is zero.

Please could you show me the solution including formula. Very much appreciated. If you can help me with my other post as well please!

Thanks!

Originally Posted by LoveDeathCab

Hey, sorry to keep asking for help but:

1. The 20th term of an arithmetic progression is 10 and the 50th term is 70.
(i) Find the first term and the common difference.

the terms of an arithmetic progression are given by the formula,
a_n = a_1 + (n - 1)d
where a_n is the nth term, a_1 is the first term, n is the current number of the term and d is the common difference.

we are told the 20th term is 10
=> a_20 = a_1 + (20 - 1)d = a_1 + 19d = 10

we are told the 50th term is 70
=> a_50 = a_1 + (50 - 1)d = a_1 + 49d = 70

so to find the first term (a_1) and the common difference (d) we must solve the system:

a_1 + 19d = 10 ....................(1)
a_1 + 49d = 70 ....................(2)

=> 30d = 60 ........................(1) - (2)
=> d = 2

but a_1 + 19d = 10
=> a_1 + 19(2) = 10
=> a_1 = -28

so the first term is -28 and the common difference is 2

(ii) Show that the su of the first 29 terms is zero.

From above, we see that the formula for our progression will be:

a_n = -28 + (n - 1)2
so a_n = -30 + 2n for n = 1,2,3,4,5...

what is the 29th term?
a_29 = -30 + 2(29) = 28

Now, the sum of the first n terms of an arithmetic progression is given by the formula,

S_n = n(a_1 + a_n)/2

so the sum of the first 29 terms is:

S_29 = 29(-28 + 28)/2 = 0

Hello, LoveDeathCab!

Hard-wire these formulas into your brain cells . . .

Given an arithmetic progression with first term a1 and common difference d,

. . [1] the nth term is: .an .= .a1 + (n-1)d

. . [2] the sum of the first n terms is: .Sn .= .½n[2·a + (n-1)d]

1. The 20th term of an arithmetic progression is 10 and the 50th term is 70.
(i) Find the first term and the common difference.

From formula [1], we have:

. . a20 .= .a1 + 19d .= .10 .(a)
. . a50 .= .a1 + 49d .= .70 .(b)


Subtract (a) from (b): .30d = 60 . → . d = 2

Substitute into (a): .a1 + 19·2 .= .10 . → . a1 = -28

(ii) Show that the sum of the first 29 terms is zero.

We have: .a1 = -28, d = 2, n = 29

Formula [2] gives us: .S29 .= .½(29) [2(-28) + 28(2)]

. . Crank it out . . .