Question

The sum of the first 9 terms of an arithmetric sequence is 81. If the sum of the first and third terms is zero, find the first term and the common difference?

Answer 1

AP
\[S_9=81...(1)\]
\[S_1+S_3=0...(2)\]
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\[S_n=\frac{n}{2}[2a+(n-1)d]\]
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For (1),
\[2a+(9-1)d=81...(3)\]
For (2)
\[2a+\cancel{(0)d}+2a+(2)d=0...(4)\]
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Simultaneous eq for (3) and (4)

Answer 2

That is what I thought but my teacher in her solutions did it with the Term Formula which makes no sense

Answer 3

So instead of Sn = n/2(2a + (n-1)d) she used Tn = a + (n-1)d

Answer 4

I think your teacher meant the sum of \(T_1\) and \(T_3\) is 0

Answer 5

Because it is a difference equation of the form tn+1 = tn+d (r=1)

Answer 6

AP
\[S_9=81...(1)\]
\[T_1+T_3=0...(2)\]
------------------------------
\[S_n=\frac{n}{2}[2a+(n-1)d]\]
------------------------------
For (1),
\[2a+(9-1)d=81...(3)\]
For (2)
\[a+a+(2)d=0...(4)\]
-----------------------------
Simultaneous eq for (3) and (4)

Answer 7

Exactly

Answer 8

That is the solution she posted with the question

Answer 9

For the second (2) equation it is \(T_1\)+ \(T_3\)=0

Answer 10

Your teacher wrote \(S_1\) and \(S_3\) = 0 which is misleading

Answer 11

Well that is what the question says. I believe she has just used the wrong formula

Answer 12

Try it out see if you get the answer

Answer 13

Thanks for the help. I will get back to studying now

Answer 14

Goodluck, ask if you have any doubts :)

Answer 15

Will do

Answer 16

The teacher is right because an arithmetic sequence would look like this:

{a, a+d, a+2d, a+3d, ... }

Hence (a) (1st term) + (a+2d) (3rd term) = 0

a=-d

Answer 17

Using the formula tn=a+(n-1)d

T9 =81 , a=-d , n=9

81= a + (9-1)d

81= (-d) + (9-1)d

81= -d +8d
7d=81

D=81/7 = 11.57

Since a=-d then a=-11.57

Answer 18

@hello @.Sam.