how many terms of series 54,51,48,....be taken so that their sum is 513? explain the double answer.
Answers
Answered by
527
use forumla ,
Sn=n/2{2a+(n-1)d}
from series ,
a=54
d=-3
Sn=513
now,
513=n/2{54 x 2+(n-1)(-3)}
1026=n{108+3-3n}
=111n-3n^2
3n^2-111n+1026=0
n^2-37n+342=0
use quadratic formula ,
n={37+_√(1369-1368)}/2
=(37+_1)/2=19 and 18
here we see n gain 2 value e.g 19 and 18
now, we check it
18th term=54+(17)(-3)=54-51=3
19th term =54-18 x 3=0
you also see 19th term is zero
so adding or no adding 19th term value of sum is always 513
so , n gain two values
Sn=n/2{2a+(n-1)d}
from series ,
a=54
d=-3
Sn=513
now,
513=n/2{54 x 2+(n-1)(-3)}
1026=n{108+3-3n}
=111n-3n^2
3n^2-111n+1026=0
n^2-37n+342=0
use quadratic formula ,
n={37+_√(1369-1368)}/2
=(37+_1)/2=19 and 18
here we see n gain 2 value e.g 19 and 18
now, we check it
18th term=54+(17)(-3)=54-51=3
19th term =54-18 x 3=0
you also see 19th term is zero
so adding or no adding 19th term value of sum is always 513
so , n gain two values
Answered by
314
Sn = n/2 ( 2a + (n-1)d)
513= n/2 (2* 54 +(n*-3) +3)
1026= 108n - 3n^2 +3n
3n^2 -111n +1026= 0
n^2 - 37n + 342 = 0
n= 18 or n = 19
the progression is towards negative side
therefore we get three answer
hopes it helped you
513= n/2 (2* 54 +(n*-3) +3)
1026= 108n - 3n^2 +3n
3n^2 -111n +1026= 0
n^2 - 37n + 342 = 0
n= 18 or n = 19
the progression is towards negative side
therefore we get three answer
hopes it helped you
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