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# help

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An arithmetic sequence with first term 1 and common difference not equal to 0 has second, tenth, and thirty-fourth terms that form a geometric sequence.

What is the thirty-fourth term of the arithmetic sequence?

The fourth term in the geometric sequence above appears as the nth term in the arithmetic sequence. Find the value of n.

Feb 4, 2019

#1
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First term =1, Common difference =1/3, Number of terms = 34. Here are the terms:

(1, 1.333333333, 1.666666667, 2, 2.333333333, 2.666666667, 3, 3.333333333, 3.666666667, 4, 4.333333333, 4.666666667, 5, 5.333333333, 5.666666667, 6, 6.333333333, 6.666666667, 7, 7.333333333, 7.666666667, 8, 8.333333333, 8.666666667, 9, 9.333333333, 9.666666667, 10, 10.33333333, 10.66666667, 11, 11.33333333, 11.66666667, 12).

The 34th term = 12

The common ratio of the geometric sequence =3

Nth term=F + D*(N - 1), where F=First term, D=Common difference, N=Number of terms

The 4th term in the geometric sequence =3 * 12 =36

36=1 + 1/3*(N - 1), solve for N

N = 106th term.

So, the 4th term of the geometric sequence, or 36, appears as 106th term in the arithmetic sequence.

Feb 4, 2019
#2
+114139
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We know that

1 + d =  a           (1)

1 + 9d = ar         (2)

1 + 33d = ar^2    (3)

Subbing (1) into (2) we have that

1 + 9d = (1 + d)r   ⇒    r =  (1 + 9d) / ( 1 + d)      (4)

Sub  (1) and  (4)  into (3)

1 + 33d  =  (1 + d)  (1 + 9d)^2 / (1 + d)^2      simplify

1 + 33d =  (1 + 9d)^2 / (1+ d)

(1 + d) (1 + 33d) = (1 + 9d)^2      simplify

33d^2 + 34d + 1 = 81d^2 + 18d + 1

48d^2 - 16d  = 0

16d ( 3d - 1) = 0

Setting both factors to 0 and solving for d  produces d = 0  (reject) or  d = 1/3  (accept)

And    1 + d = a      .....so.....1 + 1/3 = a  = 4/3

And 1 + 9d = ar

1 + 9(1/3) = (4/3)r

1 + 3 =  (4/3)r

4 = (4/3)r

r = 3

So....the 34th term   of the arithmetic series  =   1 + 33d   =   1 + 33(1/3)  = 1 + 11 =  12

Feb 4, 2019
#3
+114139
+1

The fourth term in the geometric sequence above appears as the nth term in the arithmetic sequence. Find the value of n.

The 4th term of the geometric series is   = ar^3  =   (4/3)(3)^3 = 36

So....to find which term this is in the aritmetic series, we have

36 = 1 + (1/3)(n - 1)

35 = (1/3)(n - 1)

105 = n - 1

106 = n

Feb 4, 2019