For every positive integer n, the nth term of a certain sequence is

Question

For every positive integer n, the nth term of a certain sequence is given by \((-2)^{n+2}*(\frac{6}{8^{n-1}})\). If S is the sum of the first 8 terms of the sequence, which of the following accurately represents S?

(A) \(39 > S > 36\)

(B) \(12 > S > -3\)

(C) \(\frac{-3}{4} < S< \frac{-3}{32}\)

(D) \(-32 \frac{3}{32} < S< -16 \frac{3}{4}\)

(E) \(-39 < S < -36\)

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Official Answer

E

trohera2108 · Answer

Answer E. Because the first two terms are -48 and 12 and subsequent terms will add up to a negative decimal

wadhwakaran · Answer

1st term = -48
2nd term = 12
3rd term = -3
4th term = 3/4
5th term = -3/16
6th term = 3/64
7th term = -3/256
8th term = 3/1024
The sum comes out to be somewhere between -39<S<-36

JeffTargetTestPrep · Answer

The nth term of the sequence can be written as:

(-2)^(n+2) ×   = (-2)^2 × (-2)^n ×  ] =

= 4 × (-2)^n × 48/8^n = 192 × (-2/8)^n = 192 × (-1/4)^n

We see that the sequence is a geometric sequence with a common quotient of -1/4.

The formula for the sum of the first n terms of a geometric sequence with a common quotient of q is the following:

Sum_n = a_1 × (1 - q^n)/(1 – q)

Therefore, the sum of the first 8 terms is:

Sum_8 = (-192/4) × (1 – (-1/4)^8)/(1 – (-1/4)) =

= (-192/4) × (1 – very small number)/(5/4) ≈ (-192/4) × 1/(5/4) =

= (-192/4) × (4/5) = -192/5 = -38.4

Answer: E

impeditrem · Answer

we will calculate the a1 term which will be -48, a2 = 12, a3 = -3

we will notice that r ,ie,    = -1/4 &    = -1/4

it is a geometric expression and since |r| < 1 S 8  will be closest to the sum of infinity since the addition is almost negligible.

s(infinity) =    = -192/5 =-38.4

the only option containing -38.4 is option  E ,ie,  -36 & -39

s1 = -48 
 s2 = -36 
 s3 = -39 
 S4 = -39 + 0.75 = -38.25

As n increases, the sum gets closer and closer to -38.4.

For every positive integer n, the nth term of a certain sequence is

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For every positive integer n, the nth term of a certain sequence is

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