## Trending News

# MATHS HELP please??????

show that if 19 distinct integers are chosen from the sequence 1,4,7,10,13,16,19......,97,100, there must be two of them whose sum is 104.

first of all i know the pattern is adding 3 each time but how would i determine how many numbers there will be in this pattern?

apparently this is the rule: a_n = (n-1) * k + a_1

but what does " _ " mean? it says "a_n" but i don't know what the sign in between the "a" and the "n" stand for

i mean i don't even know what those letters stand for? help?

### 3 Answers

- M3Lv 78 years agoFavorite Answer
a₁a₂a₃(ie subscripts) are not always possible on YA, so some write it as a_1 a_2 a_3, others write it as a(1) a(2) a(3) or even just a1 a2 a3 etc, denoting term 1,2,3,.....

using the notation in your q,

a_1 = 1, a_n = 100, k = 3

so 100 = 1 + (n-1)*3

(n-1)*3 = 99

n-1 = 33

n = 34

i believe i have solved the remaining part for you earlier

- 8 years ago
This sequence is in AP. Arithmetic Progression.

The common notations used is ,

first term = a

common difference = d

Tn = last term

n = number of terms.

Here a = 1 , d = 3 , Tn = 100 , n = ?

Tn = a + ( n - 1)d

100 = 1 + ( n - 1) 3

99 / 3 = n - 1

n = 34

There are 34 terms in this sequence.

But I'm not getting of how to find the earlier question.

Will try.

Seek other answers too.

Hope this helped

- 8 years ago
Let's see how many numbers there will be..(adding 3)

100:3=33(1), so there are 33+1=34 numbers

34:2=17 couples

4+100,7+97,10+94,13+91,..49+55, all of them16 couples and the two numbers 1,52

Each time when I chose 19 numbers I will find 1 couple or more