Some help on some SERIES questions please! (ASAP) (1 Viewer)

[BlackDragon]

Can someone show me how to do the following two questions?:

The sum of the first 5 terms of an arithmetic series is 35, the sum of the NEXT 5 terms is 160, find a and d.

and

Find the sum of all integers between 1 and 100 that are not multiples of 6

thanks!!
please reply asap!

 

[webby234]

S = n/2[2a + d(n-1)]

Sum (5 terms) = 35

35 = 5/2[2a + d(5-1)]
14 = 2a + 4d

Sum (10 terms) = 35 + 160 = 195

195 = 5[2a + d(10-1)]
39 = 2a + 9d

1) - 2)
25 = 5d
d = 5
a = -3

 

[webby234]

Sum of all numbers that ARE multiples of six = 6+12+18+24.....

Arithmetic series with a=6 d=6

To find no. of terms below 100

100 > a + d(n-1)
100 > 6 + 6(n-1)
n < 16.67
So there are 16 terms in the series

S = 8[12+ 6x15]
S = 816

Sum of all integers between 1 and 100
= 50 (1 +100)
= 5050

Sum of all integers not multiples of six
= 5050 - 816
= 4234

 

[BlackDragon]

hey thanks for that webby, couldn't figure out how to do that first one. cya

 

[BlackDragon]

i have a new one. please reply ASAP! anyone.
if your answer is $7041.16 don't show your working if it isn't please do.
the question is $300 invested for 5 years at 6% p.a. interest paid quaterly
thanks

 

[insert-username]

the question is $300 invested for 5 years at 6% p.a. interest paid quaterly

Assuming this is $300 invested at the start of every year:

P = 300
n = 20
r = 406/400

S = 300(406/400²⁰ + 406/400¹⁹ + 406/400¹⁸ + ... 406/400¹)

= 300 x (406/400[406/400²⁰ - 1]) ÷ (6/400)

= $7041.16.


Assuming this is $300 invested once:

P = 300
n = 20
r = 6/400

A = 300(1 + 6/400)²⁰

= $404.06


I_F