Hello, novadragon849!

Here's some help . . .

TheQ3) The 8th term of an arithmetic series is 11 and the 15th term is 39.

Find the common difference, the first term in the series and then the nth term.nth term is: .a(n) .= .a + (n-1)d

. . whereais the first term,dis the common difference.

We have: . a(8) . = -a + 7d . = .11 -[1]

. . . .and: .a(15) .= .a + 14d .= .39 .[2]

Subtract [1] from [2]: .7d = 28 . → .d = 4

Substitute into [1]: .a + 28 .= .11 . → .a = -17

Therefore: .a(n) .= .-17 + 4d

The sum of the firstQ4) Find the sum of the intergers which are divisible by 3 and lies between 1 and 400.

Then find the sum of the intergers from 1 to 400 inclusive which arenotdivisible by 3.nterms of an arithmetic series is given by:

. . . S(n) .= .½n[2a + (n-1)d]

The sum of the numbers divisible by 3 is: .3 + 6 + 9 + 12 + . . . 399

. . This is an arithmetic series with: a = 3, d = 3, n = 133

Its sum is: .S .= .½(133)[2(3) + 132(3)] .= .26,733

The sum ofallthe numbers from 1 to 400 is: .(400)(401)/2 .= .80,200

Therefore, the sum of numbernotdivisible by 3 is: .80,200 - 26,733 .= .53,467