Question

How to solve this ?

If in any decreasing arithmetic progression , sum of all its terms,except the first term is equal to -36 and the sum of all its terms, except the last term is zero and the difference of the 10th and the 6th term is equal to -16, then first term of the series is ?

Answer 1

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Answer 2

Let the first term be \(a\) and the common difference be \(d\) and the number of terms be \(n\) :)

Answer 3

ok...

Answer 4

So the series will be: \(a\), \(a+d\), \(a+2d\), ..., \(a+(n-1)d\)

Answer 5

And the sum will be \(na+\dfrac{(n-1)n}2d\)

Answer 6

Now you can set 3 equations :)

Answer 7

just a min plz ......

Answer 8

ok :)

Answer 9

so which 2 equations out of 3 should i solve ?

Answer 10

what are your 3 equations?

Answer 11

(n-1)d/2 = -36 , na = 0 , a+9d - ( a+5d ) = -16

Answer 12

so , is a 0 ?

Answer 13

well your equations are not the same as mine...

Answer 14

i am wrong anywhere ?

Answer 15

sorry for disturbing u ! ...

Answer 16

Here are my equations:
\[\left\{\begin{array}{}
na+\frac{(n-1)n}2d-a &= &-36 &\mbox{(The first term is }a) \\
na+\frac{(n-1)n}2d-[a+(n-1)d] &= &0 &\mbox{(similar to above)}\\
(a+9d)-(a+5d) &= &-16
\end{array}\right.\]

Answer 17

thanks a lot !

Answer 18

i will solve it and further get back if I have any difficulty !

Answer 19

no problem :)

Answer 20

sorry once again for disturbing u !

Answer 21

it's not disturbing me at all :)

Answer 22

gr8 of u !

Answer 23

done? :)

Answer 24

i have got d = - 4 !

Answer 25

yep, then you can find n by subtracting (2) to (1) :)

Answer 26

n=8 !

Answer 27

i have got it now !

Answer 28

Thanks a lot ! I can solve it further !

Answer 29

no problem :D
Glad to know that I've helped you :)

Answer 30

i have a question more !

Answer 31

can i post it here ? or in a new post ?

Answer 32

please open a new post :)

Answer 33

sure