A. Find the 6th term if a1=256 and r=3/2.

B. If a15=5 and r=-2 find the 20th term.

C. Find the r of a geometric sequence if a1 =4 and the 10th term is 78125000

Question

PRECIOUSRECORTE47IN PRECIOUSRECORTE47IN

Math

Answered

A. Find the 6th term if a1=256 and r=3/2.

B. If a15=5 and r=-2 find the 20th term.

C. Find the r of a geometric sequence if a1 =4 and the 10th term is 78125000

Sagot :

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RHOZETAIN RHOZETAIN · Answer 1

✏️ Geometric Sequence

[tex] {\Large{\overline{\underline{\sf{\hookrightarrow Answers:}}}}} [/tex]

[tex] \sf a_6 = 1944 [/tex]
[tex] \sf a_{20} = -160 [/tex]
[tex] \sf r = 5 [/tex]

Solution:

Here, we use the formula for the general term in a geometric sequence:

[tex] {\Large{\boxed{\sf{a_n = a_1 r^{n-1}}}}} [/tex]

A.

Given that:

the first term [tex] \sf a_1 [/tex] = 256
the common ratio [tex] \sf r [/tex] = [tex] \sf \frac{3}{2} [/tex]
the number of terms [tex] \sf n [/tex] = 6

Solve:

[tex] \sf{a_n = a_1 r^{n-1}} [/tex]
[tex] \sf{a_6 = (256)(\frac{3}{2})^{6-1}} [/tex]
[tex] \sf{a_6 = (256)(\frac{3}{2})^{5}} [/tex]
[tex] \sf{a_6 = (256)(\frac{243}{32})} [/tex]
[tex] {\sf \therefore a_6 = {\boxed{\green{\sf{1944}}}}} [/tex]

B.

Solution 1:

Given that:

the fifteenth term [tex] \sf a_{15} [/tex] = 5
the common ratio [tex] \sf r [/tex] = -2
the number of terms [tex] \sf n [/tex] = 15

Solve for the first term.

[tex] \sf{a_n = a_1 r^{n-1}} [/tex]
[tex] \sf{5 = a_1 (-2)^{15-1}} [/tex]
[tex] \sf{5 = a_1 (-2)^{14}} [/tex]
[tex] \sf{5 = a_1 (16384)} [/tex]
[tex] \sf{a_1 = \frac{5}{16384}} [/tex]

Solve for [tex] \sf a_{20} [/tex]. Let [tex] \sf n [/tex] = 20

[tex] \sf{a_n = a_1 r^{n-1}} [/tex]
[tex] \sf{a_{20} = (\frac{5}{16384}) (-2)^{20-1}} [/tex]
[tex] \sf{a_{20} = (\frac{5}{16384}) (-2)^{19}} [/tex]
[tex] \sf{a_{20} = (\frac{5}{16384}) (-524288)} [/tex]
[tex] \sf{a_{20} = \frac{-2621440}{16384}} [/tex]
[tex] {\sf \therefore a_{20} = {\boxed{\green{\sf{-160}}}}} [/tex]

Solution 2:

There are six terms between [tex] \sf a_{15} [/tex] and [tex] \sf a_{20} [/tex]. So we let [tex] \sf n [/tex] = 6

Solve:

[tex] \sf{a_{20} = a_{15} r^{6-1}} [/tex]
[tex] \sf{a_{20} = (5) (-2)^{5}} [/tex]
[tex] \sf{a_{20} = (5) (-32)} [/tex]
[tex] {\sf \therefore a_{20} = {\boxed{\green{\sf{-160}}}}} [/tex]

C.

Given that:

the first term [tex] \sf a_1 [/tex] = 4
the tenth term [tex] \sf a_{10} [/tex] = 7812500
the number of terms [tex] \sf n [/tex] = 10

Solve:

[tex] \sf{a_n = a_1 r^{n-1}} [/tex]
[tex] \sf{7812500 = (4) r^{10-1}} [/tex]
[tex] \sf{7812500 = (4) r^{9}} [/tex]
[tex] \sf{r^{9} = \frac{7812500}{4}} [/tex]
[tex] \sf{r^{9} = 1953125} [/tex]
[tex] \sf{r = \sqrt [9] {1953125}} [/tex]
[tex] {\sf \therefore r = {\boxed{\green{\sf{5}}}}} [/tex]

[tex]{\: \:}[/tex]

[tex] {\huge{\overline{\sf{Hope\:It\:Helps}}}} [/tex]

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