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# sequences question

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1. the question is, The first term of a arithmetic sequence is six times the 5th term and the sum of the 2nd and 8th term is 8, find the 20th term of this sequence.

Im kinda confused, I know i need to find A and D and can get a one of the simultaneus equations, but the 2nd one reguarding the sum of the 2nd and 8th term is is confusing to make a simultaneus equation?
2. (Original post by SunDun111)
the question is, The first term of a arithmetic sequence is six times the 5th term and the sum of the 2nd and 8th term is 8, find the 20th term of this sequence.

Im kinda confused, I know i need to find A and D and can get a one of the simultaneus equations, but the 2nd one reguarding the sum of the 2nd and 8th term is is confusing to make a simultaneus equation?
2nd term is A + d

8th term is A + 7d

Sum of 2nd and 8th term is A + d + A + 7d = 8
3. and the 5th term is A + 4d ?
4. (Original post by Zacken)
2nd term is A + d

8th term is A + 7d

Sum of 2nd and 8th term is A + d + A + 7d = 8
(Original post by SunDun111)
the question is, The first term of a arithmetic sequence is six times the 5th term and the sum of the 2nd and 8th term is 8, find the 20th term of this sequence.

Im kinda confused, I know i need to find A and D and can get a one of the simultaneus equations, but the 2nd one reguarding the sum of the 2nd and 8th term is is confusing to make a simultaneus equation?
thanks guys
5. (Original post by the bear)
and the 5th term is A + 4d ?
i already worked out the equation, got the answer right but the 5th term is

5A + 24D = 0 when you figure out properly
6. (Original post by SunDun111)
thanks guys
No problem.
7. (Original post by Zacken)
No problem.
came acorss another question, it is The sum of the first 30 terms of an arthimetic series is 90, The sum of the next 10 terms is also 90, Find the first 1st term and common difference,

i thought i'd use the sum formula, and got
1/2 multiplied by 30 (2a + 29d) = 90
and my second one would be

1/2 multiplied by 40(2a + 39d) = 90
i cant get the right answer have i made my equations wrong
8. (Original post by SunDun111)
1/2 multiplied by 40(2a + 39d) = 90
i cant get the right answer have i made my equations wrong
Yes, this bit is wrong. It says the sum of the next 10 terms is 90.

That is, the 31st + 32nd + 33rd + ... + 40th term is 90.

i.e: A + 30d + A + 31d + A + 32d + ... + A + 39d = 10A + 345d = 90.
9. (Original post by Zacken)
Yes, this bit is wrong. It says the sum of the next 10 terms is 90.

That is, the 31st + 32nd + 33rd + ... + 40th term is 90.

i.e: A + 30d + A + 31d + A + 32d + ... + A + 39d = 10A + 345d = 90.
(Original post by Zacken)
Yes, this bit is wrong. It says the sum of the next 10 terms is 90.

That is, the 31st + 32nd + 33rd + ... + 40th term is 90.

i.e: A + 30d + A + 31d + A + 32d + ... + A + 39d = 10A + 345d = 90.
not really following here, wouldnt it be 0.5 x 10 ( 2a + 9D ) = 90?
10. (Original post by SunDun111)
not really following here, wouldnt it be 0.5 x 10 ( 2a + 9D ) = 90?
No, that's the first 10 terms.
11. (Original post by Zacken)
No, that's the first 10 terms.
yeah but from the 30th to 40th they is 10 terms, so wouldnt it still be 0.5 multipled by 10? even though your right what im kinda confused about how you got the second equation
12. (Original post by SunDun111)
yeah but from the 30th to 40th they is 10 terms, so wouldnt it still be 0.5 multipled by 10? even though your right what im kinda confused about how you got the second equation
No. Your formula only works if it is the first ten terms. Not any random set of ten terms.

My equation is just:
31st term: A + 30d
32nd term: A + 31d
.
.
.
40th term: A + 39d

A + 30d + A + 31d + ... + A + 39d
That's 10A + 345d
13. (Original post by Zacken)
No. Your formula only works if it is the first ten terms. Not any random set of ten terms.

My equation is just:
31st term: A + 30d
32nd term: A + 31d
.
.
.
40th term: A + 39d

A + 30d + A + 31d + ... + A + 39d
That's 10A + 345d
Ok thanks i understand
14. (Original post by SunDun111)
Ok thanks i understand
Awesome.

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