Show work and explain with formulas if possible.

12. You enlarge a photo to 150% of its previous size several times. If the photo measures 4 inches in length after the first increase, how long is it after 5 increases?

13. Larry exercised every day for 30 days. The first day he did 8 sit-ups, the 2nd day he did 12 sit-ups, the 3rd day he did 16 sit-ups, etc. After the 30th day of exercising, what was total number of sit-ups that Larry had done?

16. Write the first three terms of the series and then evaluate the sum of the series. (See picture)

[tex]a_1=4,\ r=1.5,\ n=5\\\\a_n=a_1\cdot r^{n-1}\\\\a_5=4\cdot (1.5)^{5-1}\\\\.\ =4\cdot (1.5)^4\\\\.\ =4\cdot 5.0625\\\\.\ =\large\boxed{20.25}[/tex]

13 Answer: 1980

Step-by-step explanation:

[tex]a_1=8,\ d=4, \n=30\\\\a_n=a_1+d(n-1)\\\\a_{30}=8+4(30-1)\\\\.\ =8+4(29)\\\\.\ =8+116\\\\.\ =124\\\\\\S_n=\dfrac{a_1+a_n}{2}\cdot n\\\\\\S_{30}=\dfrac{8+124}{2}\cdot 30\\\\\\.\quad =\dfrac{132}{2}\cdot 30\\\\\\.\quad =66\cdot 30\\\\\\.\quad =\large\boxed{1980}[/tex]

16 Answer: 12

Step-by-step explanation:

[tex]\sum\limits^\infty_{n=1} 6\bigg(\dfrac{1}{2}\bigg)^{n-1}\\\\\\S_1=6\qquad \qquad S_2=3\qquad \qquad S_3=1.5\\\\\\\sum\limits^\infty_{n=1} 6\bigg(\dfrac{2}{2^n}\bigg)=\sum\limits^\infty_{n=1} 6\cdot 2\bigg(\dfrac{1}{2^n}\bigg)=12+\sum\limits^\infty_{n=1}\bigg(\dfrac{1}{2^n}\bigg)=12+0=\large\boxed{12}[/tex]

nbayoungboy0421 nbayoungboy0421 · Answer 2 · 2019-04-16T21:18:20+02:00

12 Answer: 20.25

Step-by-step explanation:

\begin{lgathered}a_1=4,\ r=1.5,\ n=5\\\\a_n=a_1\cdot r^{n-1}\\\\a_5=4\cdot (1.5)^{5-1}\\\\.\ =4\cdot (1.5)^4\\\\.\ =4\cdot 5.0625\\\\.\ =\large\boxed{20.25}\end{lgathered}

a

1

=4, r=1.5, n=5

a

n

=a

1

⋅r

n−1

a

5

=4⋅(1.5)

5−1

. =4⋅(1.5)

4

. =4⋅5.0625

. =

20.25

13 Answer: 1980

Step-by-step explanation:

16 Answer: 12

Step-by-step explanation:

\begin{lgathered}\sum\limits^\infty_{n=1} 6\bigg(\dfrac{1}{2}\bigg)^{n-1}\\\\\\S_1=6\qquad \qquad S_2=3\qquad \qquad S_3=1.5\\\\\\\sum\limits^\infty_{n=1} 6\bigg(\dfrac{2}{2^n}\bigg)=\sum\limits^\infty_{n=1} 6\cdot 2\bigg(\dfrac{1}{2^n}\bigg)=12+\sum\limits^\infty_{n=1}\bigg(\dfrac{1}{2^n}\bigg)=12+0=\large\boxed{12}\end{lgathered}

n=1

∑

∞

6(

2

1

)

n−1

S

1

=6S

2

=3S

3

=1.5

n=1

∑

∞

6(

2

n

2

)=

n=1

∑

∞

6⋅2(

2

n

1

)=12+

n=1

∑

∞

(

2

n

1

)=12+0=

12