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5 votes
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)



Show work and explain with formulas if possible. 12. You enlarge a photo to 150% of-example-1

2 Answers

4 votes

12 Answer: 20.25

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

Explanation:

16 Answer: 12

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

User Ptamzz
by
5.8k points
2 votes

12 Answer: 20.25

Explanation:


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}

13 Answer: 1980

Explanation:


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

16 Answer: 12

Explanation:


\sum\limits^\infty_(n=1) 6\bigg((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((2)/(2^n)\bigg)=\sum\limits^\infty_(n=1) 6\cdot 2\bigg((1)/(2^n)\bigg)=12+\sum\limits^\infty_(n=1)\bigg((1)/(2^n)\bigg)=12+0=\large\boxed{12}