a+n-1{D} the value of 'n' has positive or negative  values, if it gets negative values question will be wrong. for this question iam getting  n= -25 .but ans shows n=27. negative ans i got is wrong how ?explain clarly
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in arithmatic progression the value of (n) can be negative . The 26th, 11th and the last term of an A.P. are 0, 3 and -1/5 respectively. Find the common difference and the number of terms.

The expresion for the nth term of an arithmetic progression is given by
a_n = a_1 + (n - 1)d

Given: a_26 = 0, a_11 = 3
a_26 = 0 therefore 0 = a_1 + 25d    since a_26 = a_1 + (26 – 1)*d
a_11 = 3, therefore 3 = a_1 + 10d    since a_11 = a_1 + (11 – 1)*d
Subtracting the 2nd eqn from the 1st eqn,


-3 = 15d
d = -1/5,  and a_1 = 5


Let last term in the series be a_n, then using
a_n = a_1 + (n - 1)d, we get


-1/5 = 5 + (n - 1)*(-1/5)
1 = -25 + (n - 1)
n = 27

 

by Level 11 User (81.5k points)
edited by
thanks for helping me ,but what are slashes indicating can u edit it please for me ,and also explain me the whole question
Hi there. What slashes do you mean? I don't see any. Do you mean the underscores, as in a_1, a_2, a_3, etc?

These just indicate the successive terms in the sequence. a_1 is the 1st term, a_2 is the 2nd term, and so on.

Does that make the solution clearer?
yes ur right they are undercores ,and my next question is about the ans n=- 25 actually ans is n=27.is it correct, if not why can u explain about it
OK, the ans isn’t n = -25. That’s wrong, I’m afraid. The correct ans is n = 27. If you got n = -25, then it looks like you made a mistake in your calculations and ended up using the wrong equations. I would need to see your working to be certain.
You gave an expression “a+n-1{D}”, which is similar to my expression, a_n = a_1 + (n – 1)d, but not quite right. If you used that expression then it may explain why you get n = -25.
HTH
hi

actually i want to know one thing in this calculations about reciprocals  i'e.about -1/5,         i have done this problem using reciprocals ,when its side changed,it will be -5/1 or only just 5/1.give me quick reply.this is where i am confused in getting the correct answer
Hi there,
I’m afraid I’m getting a bit confused myself. The problem I’m having is that I don’t know what work you have done, so I’m not too sure how I can figure out where you may have gone wrong.
There is also the problem of language. You mention reciprocals, but -1/5 is actually a fraction, cf slash/underscore. This makes it difficult for me to be certain about what you’re saying.
You already have been given two solutions to the problem by two different persons. I’m guessing that you want to know where you went wrong in your own solution.
I think the best, and fastest solution, would be for you to upload your working of this problem, and then I can point out what went wrong.

hi,

feel free ! and don't get frightened and be cool , everything will go on smoothly. 

now here i am going to show my work to u ,now take a look. let's take the last part of the problem where i am confused

an=a+(n-1)d

-1/5=5+(n-1)-1/5

-1/5-5=(n-1)-1/5

-26/5=(n-1)-1/5

at this particular point iam confused how the reciprocal is taken here.if i will change the side  (-1/5) ,it will become -5/1 or simply 5/1.this point needs to be clarified.i have done it like this

-26/5*5/1=(n-1)

-130/5=(n-1)

-26+1=n

n=-25

OK, seenvas, I see where the problem is now.

You have
-26/5=(n-1)-1/5

And you want to simplify this.
So multiply both sides by the same value, which in this case will be (-5/1)
Multiplying both sides by (-5/1) gives us,

-26/5*(-5/1)=(n-1)(-1/5)*(-5/1)
Cancelling out the minus signs gives us

26/5*(5/1)=(n-1)(1/5)*(5/1)

Which simplifies to,
26 = (n – 1)   --> n = 27

You made a mistake earlier, in the two lines
-26/5*5/1=(n-1)  ----- both sides of this eqn should be the same sign
-130/5=(n-1)        ------  this should be 130/5 = n-1, or 26 = n - 1

hi

well that's fine.i got one point from ur explanation i'e . the sign of a reciprocal has not changed.i mean to say that fraction -1/5. this is the  major key  point for me .becoz this is where iam struggling.

 fermat are u a proffessional maths teacher.thanks for helping me .keep helping me
Hi Seenvas,
No. I'm not a professional maths teacher. I am simply a retired and rather elderly gentleman with a penchant for maths. I also have a masters degree (M.Sc.) in maths, from the open university. In addition to that I have had many years experience of going to forums online and answering questions on maths, dynamics, kinematics, mechanics and some physics. (But not statistics!!). As I said, I am now retired and consider forums like this as an occasional pastime rather than a fulltime occupation.
If you would like to get good at maths, then do lots of practice. Start with simple problems and solve them. Look at some of the problems on here and see how they are solved. Rod is the number one user on here. Have a look at some of his solutions. The key is in understanding what you're doing, how you solved the problem. That is more important than simply remembering how to go through some mathematical process (but that's good too :) ) When you understand how simple(r) problems are solved, then look at more difficult ones.
Enjoy learning maths!

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