Anonymous
Anonymous asked in Science & MathematicsMathematics · 8 years ago

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

Relevance
  • M3
    Lv 7
    8 years ago
    Favorite 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

Still have questions? Get your answers by asking now.