Question

How many terms of the series 54,51,58 be taken to the sum 513 explain the double answer

Answer 1

Answer:

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Step-by-step explanation:

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

Hope it helps u :-)

Answer 2

Answer:

Step-by-step explanation:

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

3n^2-111n+1026=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