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

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.

9) i can't see the angle so whatever it is just do this: cos(angle) x hypotenuse and then use that answer and they hypotenuse and plug it into this: (answer you just got)^2 + b^2 = (hypotenuse)^2. the first answer will be the answer of the adjacent side and the second answer after this equation will be the opposite side.

10) y = 14 x = √147 or 12.12435565

11) x = 4.8

12) x = 4.2

13) x = 9.3

14) pq: 13.2

qr : 17.6

angle p: 53º

15) angle c: 54.3º

angle a: 35.7º

16) h = 21040

v = 1841

step-by-step explanation:

10) 7/sin(30) = 14 = y

7^2 + b^2 = 14^2

49 + b^2 = 196

b^2 = 196 - 49

√b^2 = √147

x = √147

11) cos(70) x 14 = 4.788282007

x = 4.8

12) tan(25) x 9 = 4.196768923

x = 4.2

13) sin(31) x 18 = 9.270685348

x = 9.3

14) pq: sin(37) x 22 = 13.23993051

pq: 13.2

qr: cos(37) x 22 = 17.56998122

qr : 17.6

angle p: 180 - (37 + 90) = 53º

15) cos-1(7/12) = 54.31466529

angle c: 54.3º

angle a: 180 - (90 + 54.31466529) = 35.68533471

angle a: 35.7º

16) 4 x 5280 = 21120

cos(5) x 21120

h = 21039.63202

h = 21040

sin(5) x 21120

v = 1840.729287

v = 1841