Question

How many terms of the series 54,51,48,..., be taken so that their sum is
513?Explain the double answer.​

Answer 1

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

Answer 2

Given :

• First term ( a ) = 54

• Common difference ( d ) = - 3

• Sum of terms ( Sn ) = 513

To Find :

• No. of terms in given A.P

Solution :

Sn = n/2 [ 2a + ( n - 1 ) d

513 = n/2 [ 2 × 54 + ( n - 1 ) ( - 3 )

1026 = n [ 108 - 3n + 3 ]

1026 = 108n - 3n² + 3n

3n² - 111n + 1026 = 0

n² - 37n + 342 = 0

n² - 19n - 18n + 342 = 0

n ( n - 19 ) - 18 ( n - 19 ) = 0

( n - 19 ) ( n - 18 ) = 0

n = 19

n = 18

Here n has two values . So let's which is correct

S₁₉ = 19/2 [ 2 × 54 + 18 × ( - 3 )

S₁₉ = 19/2 [ 108 - 54 )

S₁₉ = 19/2 [ 54 ]

S₁₉ = 19 × 27

S₁₉ = 513

Given A.P has 19 terms