How many terms of series 51 41 48 be taken so that their sum is 513?
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given
a = 54
d = -3
sn = 513
sn = n/2(2a+(n-1)d)
513 = n/2(108+3-3n)
1026=111n-3n^2
n^2-37n + 342 = 0
on solving you will get
n=18, n=19
you are getting here two values, both positive so, both are applicable
since, d=negative
a case will come where the number after 0 will cancel out the number before 0,
positive number will cancel negative number
so, both the cases are applicable
you may try for both taking n=18 and n=19, you will get the same answer.
I hope it helps you. It is not the solution of your question instead, i took an example, just to help you understand.
All the best!!! Thumbs up!
a = 54
d = -3
sn = 513
sn = n/2(2a+(n-1)d)
513 = n/2(108+3-3n)
1026=111n-3n^2
n^2-37n + 342 = 0
on solving you will get
n=18, n=19
you are getting here two values, both positive so, both are applicable
since, d=negative
a case will come where the number after 0 will cancel out the number before 0,
positive number will cancel negative number
so, both the cases are applicable
you may try for both taking n=18 and n=19, you will get the same answer.
I hope it helps you. It is not the solution of your question instead, i took an example, just to help you understand.
All the best!!! Thumbs up!
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