Sum of first four terms of a geometric series is #15# and next four terms is #240#. Find the first term and common ratio of the series?

Answer 1

There are two possibilities i.e. either #a_1=1# and #r=2#

or #a_1=-3# and #r=-2#.

In a geometric sequence, whose first term is #a_1# and common ratio is #r#, while #n^(th)# term #a_n=a_ar^((n-1))#, sum of first #n# terms is given by #S_n=(a_1(r^n-1))/(r-1)#.
Here assume that first term is #a_1# and common ratio is #r#. As sum of first four terms is #15#
#S_4=(a_1(r^4-1))/(r-1)=15# .........................(A)
i.e. #a_1+a_1r+a_1r^2+a_1r^3=15#
i.e. #a_1(1+r+r^2+r^3)=15# .........................(1)
and #a_1(r^4+r^5+r^6+r^7)=240#
i.e. #a_1r^4(1+r+r^2+r^3)=240# .........................(2)

Dividing (2) be (1), we get

#r^4=240/15=16#
i.e. #r=+-2#
If #r=2#, #a_1=15/(1+2+4+8)=1#
and if #r=-2#, #a_1=15/(1-2+4-8)=15/(-5)=-3#.

Hence sequence is

either #{1,2,4,8,16,32,64,18}# i.e. #a_1=1# and #r=2#
or #{-3,+6,-12,+24,-48,96,-192,384}# i.e. #a_1=-3# and #r=-2#
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Answer 2

Let the first term of the geometric series be (a) and the common ratio be (r).

Given that the sum of the first four terms is 15, we have:

[a + ar + ar^2 + ar^3 = 15]

And the sum of the next four terms is 240, so:

[ar^4 + ar^5 + ar^6 + ar^7 = 240]

Using the formula for the sum of a geometric series, we can express these sums in terms of (a) and (r):

[S_4 = a\left(\frac{{r^4 - 1}}{{r - 1}}\right)] [S_8 = a\left(\frac{{r^8 - 1}}{{r - 1}}\right)]

Given (S_4 = 15) and (S_8 = 240), we can set up the following equations:

[a\left(\frac{{r^4 - 1}}{{r - 1}}\right) = 15] [a\left(\frac{{r^8 - 1}}{{r - 1}}\right) = 240]

Divide the second equation by the first equation:

[\frac{{\frac{{r^8 - 1}}{{r - 1}}}}{{\frac{{r^4 - 1}}{{r - 1}}}} = \frac{{240}}{{15}}]

Simplify:

[\frac{{r^8 - 1}}{{r^4 - 1}} = 16]

Multiply both sides by (r^4 - 1):

[r^8 - 1 = 16(r^4 - 1)]

Expand:

[r^8 - 1 = 16r^4 - 16]

Move terms to one side:

[r^8 - 16r^4 - 1 + 16 = 0]

Factor as a quadratic in (r^4):

[(r^4)^2 - 16r^4 + 15 = 0]

Let (u = r^4), then:

[u^2 - 16u + 15 = 0]

Solve this quadratic equation for (u):

[(u - 1)(u - 15) = 0]

So, (u = 1) or (u = 15).

If (u = 1), then (r^4 = 1), and (r = 1) or (r = -1).

If (u = 15), then (r^4 = 15), and (r) is a complex number.

Since it's unlikely for a geometric series to have a common ratio of -1, we'll discard that solution. Thus, (r = 1).

Now, substitute (r = 1) into one of the original equations:

[a + a + a + a = 15]

[4a = 15]

[a = \frac{15}{4} = 3.75]

So, the first term of the series is 3.75 and the common ratio is 1.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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