Thread: Arithmetic series..... other problems..

An arithmetic series has first term a and common difference d.
The sum of the first 31 terms of the series is 310.

a) Show that a + 15d = 10

> S31 = 31/2 ( 2a + (31-1) d)
> 31( a + 15d ) = 310
> a + 15d = 310/31; a + 15d = 10

b)Given also that the 21st term is twise the 16th term, find the value of d.

???

c)the nth term of the series is un. Given that
K
Sum un = 0, find the value of K.
n = 1

???

Originally Posted by ansonbound

An arithmetic series has first term a and common difference d.
The sum of the first 31 terms of the series is 310.

a) Show that a + 15d = 10

> S31 = 31/2 ( 2a + (31-1) d)
> 31( a + 15d ) = 310
> a + 15d = 310/31; a + 15d = 10

b)Given also that the 21st term is twise the 16th term, find the value of d.

$\displaystyle 2(a+20d)=a+15d$

so

$\displaystyle a+25d=0$

and you already have:

$\displaystyle a+15d=10$

So solve these simultaneous equations.

CB

Originally Posted by ansonbound

c)the nth term of the series is un. Given that
K
Sum un = 0, find the value of K.
n = 1

???

From the previous part you will now know $\displaystyle a$ and $\displaystyle b$, so plug them into the formula for the sum, which gives a equation in K to solve:

$\displaystyle \sum_{n=1}^K u_n=\frac{K}{2}(2a+(K-1)d)=0$

CB