easy maths question :( (1 Viewer)

[coroneos]

I am feeling really stupid at the moment because I can't figure out the correct way to solve the problem.

Find the sum of all integers between 20 and 50 that are divisible by 3.

This question is under the topic of Sequences and series. Now, how do I go about doing it?

 

[Sammmmmuel]

find the first term that is divisible by 3... it will be 21 and thats ur first term(a). now find out how many numbers between 20 and 50 are divisible by 3 and that will be the number of terms in the series (n). the common difference (d) is 3. Now use the equation for sum of n terms of an arithmetic series... it think thats how u do it?

 

[Huy]

T1 = 21
T(n) = 48
n = 16

S(16) = 16/2 [21+48]
= 8(69)
= 552

 

[Lexicographer]

Couldn't have said it better.

 

[Fply]

T(0)=21 (initial conditon)
T(n)= 48 - 21
n =9 (this means since T(0) the ninth term is T(9) otherwords the total numbers of terms that could divisible by 3 is 10)

the sum of those No. would be (10)*(48+21)/2
=345

 

[fatmuscle]

if you're really desparate, do it by hand!!!

 

[fatmuscle]

i'm assuming you're checking your work...

don't do it in an actual exam

 

[Kurby]

a = 21 (first term)
L = 48 (last term)
d = 3

Find n first..
Tn = a + (n-1)d
48 = 21 + (n-1)3
48 = 21 + 3n - 3
10 = 3n
n = 10

using sum formular.

Sn = n/2 x (a+L)
Sn = 10/2 (21+48)
Sn = 5 (69)
Sn = 345!!

There you go. Full setting out.

 

[Fply]

I like this solution, it is more clear

 

[pseudo]

use an addition reaction starting at 21 with additions of 3