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1/2, 3/4, 1, …What are the 6th and 7th terms of the sequence?

User Virmundi
by
6.5k points

1 Answer

7 votes

Answer:

Answer

It can be (1+x)

n

only as (1−x)

n

terms are +ve/−ve alternatively hence can't be in A.P.,

General term, T

rH

=

n

C

r

⋅x

r

⇒T

5

=

n

C

4

⋅x

4

,T

6

=

n

C

5

⋅x

5

,T

7

=

n

C

6

⋅x

6

Coefficients are in A.P.

⇒2

n

C

5

=

n

C

4

+

n

C

6

5!(n−5)!

2n!

=

4!(n−4)!

n!

+

6!(n−6)!

n!

5(n−5)

2

=

(n−4)(n−5)

1

+

6×5

1

⇒2(n−4)=5+

6

1

(n−4)(n−5)

⇒12(n−4)=30+n

2

−9n+20

12n−48=n

2

−9n+50

⇒n

n

−21n+98=0

⇒n

2

−14n−7n+98=0

⇒n(n−14)−7(n−14)=0

⇒(n−7)(n−14)=0

n=7 or n=14.

User Amitsharma
by
7.3k points
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