Arithmetic Sequence help

This is on one of the gcse practice papers but I don't know what to do, it would really be appreciated if somebody helped me go through this with me:
"The nth term of an arithmetic sequence is given by an+b where a and b are integers. The 5th term is 19 and the 11th term is 42. Work out the values of a and b."

Reply 1

EnglishNoobC

12

Simultaneous equations

EDIT: 5a+b
11a+b

Posted from TSR Mobile

(edited 6 years ago)

Reply 2

Mermaidqueen

OP

14

Original post by cotton candie

That looks right but it can't be because the question states that theyre integers

Reply 3

Muttley79

20

Original post by cotton candie

...

Please remove your full solution.

PS it's not well explained btw as you've used 'n' to mean two different things.

I think the 11th term should be 43 to get integer solutions.

(edited 6 years ago)

Reply 4

thekidwhogames

18

Un = nth term

Un = a + (n-1)d ; this is the nth term of an arithmetic sequence

Given U5 = 19 and U11 = 42

Therefore, U5 = a + 4d and U11 = a + 10d

Therefore a + 4d = 19 and + 10d = 42

Therefore 6d = 23 therefore 23/6 = d

Solving for a = 3.666... = 3 + 2/3 = 11/3

a = 1st term, d = common difference of an arithmetic sequence

Therefore, Un = a + (n-1)d

You cannot have integer solutions. This is impossible. Please check the question and/or mark scheme.

Reply 5

Mermaidqueen

OP

14

QUOTE=thekidwhogames;71729076]Un = nth term

Un = a + (n-1)d ; this is the nth term of an arithmetic sequence

Given U5 = 19 and U11 = 42

Therefore, U5 = a + 4d and U11 = a + 10d

Therefore a + 4d = 19 and + 10d = 42

Therefore 6d = 23 therefore 23/6 = d

Solving for a = 3.666... = 3 + 2/3 = 11/3

a = 1st term, d = common difference of an arithmetic sequence

Therefore, Un = a + (n-1)d

You cannot have integer solutions. This is impossible. Please check the question and/or mark scheme.

This is the mark scheme but I'm really confused: (its question 2 by the way)

Reply 6

zeldor711

18

Original post by Mermaidqueen

This is on one of the gcse practice papers but I don't know what to do, it would really be appreciated if somebody helped me go through this with me:
"The nth term of an arithmetic sequence is given by an+b where a and b are integers. The 5th term is 19 and the 11th term is 42. Work out the values of a and b."

Looking at the mark scheme you posted it seems that the 11th term is 43, not 42. Therefore just form:
5a + b = 19
11a + b = 43

Subtract to eliminate b, solve for a and sub back in for b

Reply 7

thekidwhogames

18

Original post by Mermaidqueen

QUOTE=thekidwhogames;71729076]Un = nth term

Un = a + (n-1)d ; this is the nth term of an arithmetic sequence

Given U5 = 19 and U11 = 42

Therefore, U5 = a + 4d and U11 = a + 10d

Therefore a + 4d = 19 and + 10d = 42

Therefore 6d = 23 therefore 23/6 = d

Solving for a = 3.666... = 3 + 2/3 = 11/3

a = 1st term, d = common difference of an arithmetic sequence

Therefore, Un = a + (n-1)d

You cannot have integer solutions. This is impossible. Please check the question and/or mark scheme.

This is the mark scheme but I'm really confused: (its question 2 by the way)

Oh but it's different to the number you said

Reply 8

Scott_/11

What mark scheme is this

Reply 9

Rebel1786

1

where did you get the mark scheme

Reply 10

Rebel1786

1

Original post by Mermaidqueen

QUOTE=thekidwhogames;71729076]Un = nth term

Un = a + (n-1)d ; this is the nth term of an arithmetic sequence

Given U5 = 19 and U11 = 42

Therefore, U5 = a + 4d and U11 = a + 10d

Therefore a + 4d = 19 and + 10d = 42

Therefore 6d = 23 therefore 23/6 = d

Solving for a = 3.666... = 3 + 2/3 = 11/3

a = 1st term, d = common difference of an arithmetic sequence

Therefore, Un = a + (n-1)d

You cannot have integer solutions. This is impossible. Please check the question and/or mark scheme.