Exercise Math

Exercise.

1.       The 1^st term of arithmetic sequences is 6, and the 5^rd term is 22. Find the value of the divergent of the arithmetic sequences. (Chapter 5 pages 195)

2.       In 5 weeks, Budi has been training to face the marathon competition. Each week he has to run twice time further than the week before. In the 3^rd week he runs 3 km. determine the total distance for Budi in five weeks training. (Chapter 5 pages 213)

3.       L et the arithmetic series(t+23)+(t+17)+(t+11)+… 

a.       Find out the value of the divergent.

b.      Find out the value of U₅ and U₆.

c.       Calculate the first six terms in series above.

(Chapter 5 pages 201)

4.       The first round in tennis table competition was followed by 128 teams. The second round was followed by 64 teams , the third round was followed by 32 teams, and so on. In what round will the competition reach the final (only followed by 2 teams)? (Chapter 5 pages 199)

5.       (Chapter 5 pages 189)

a.       Copy the figure of Pascal triangle and then continue until the 10^th line.

b.      Find the sum of following Pascal triangle numbers lines.

(i)                  The 8^th lines.

(ii)                The 9^th lines.

(iii)               The 10^th lines.

(iv)              The 11^th lines

(v)                The 12^th lines.

Solution.

1.       We know, the 1^st term (a) = 6 and The 5^th term (U₅ )= 22

For  U_n=a+(n-1)b, then

U₅=a+(n-1)b

U5=a+4b

From the formula, we can find the value of the divergent of the arithmetic sequences.

22=6+4b

4b=22-6

b=4

So, the divergent value is 4.

2.       From the question we know it is geometric series. So, we must use the formula of geometric series.

We know, n = 5, r = 2, and U3 = 4

Formulas of Geometric Series.

U_n=ar^n-1  

S_n=a[(1-rⁿ)/(1-r)]

First, we must find the value of 1^st term (a).

U₃=4

4=a 4

a=1

Next, we can determine the total distance (S₅)

S_n=a[(1-rⁿ)/(1-r)]

S_n=1[(1-2⁵)/(1-2)]

S₅ =1(-31/-1)

S₅ =31

So, the total distance for Budi in five weeks trining is 31 km.

3.       (t+23)+(t+17)+(t+11)+…  is the arithmetic series.

a.       Find the divergent

Un=a+(n-1)b

We know, U2=(t+17) and a=(t+23)

t+17=t+23+(2-1)b

b=-6

So, the divergent from the arithmetic series above is -6.

b.      Find U₅ and U₆

U_n=a+(n-1)b

U₅=(t+23)+(5-1)(-6)

U₅=(t-1)

So, the value of the 5^th term is (t-1)

U₆=U₅+b

U₆=t-7

So, the value of the 6^th term is (t-7)

c.       Calculate the first six terms in series above.

S_n=n/2(a+U_n)

S6=6/2[(t+23)+(t-7)]

S6=3(2t+16)

S6=(6t+48)

Calculate the first six terms is (6t+48)

4.       From the question we know it is Geometric Sequences. So, we must use the formula of geometric sequences. The formula of geometric sequences for the n-term is given as follow.

U₅=a r^n-1

We know, a = 128, U₂=64, U₃=32,and U_n=2

Before find the value of n, we need ratio(r)

r=U₂/U₁=1/2

Next, we can find the value of n.

U_n=a r^n-1

2=128(1/2)^n-1

(1/2)^n-1=(1/2)⁶

So,

n-1=6

n=7

So, the competition will reach the final in 7^th round.

5.       Pascal Triangle

The 1st line The 2nd line The 3rd line The 4th line The 5th line The 6th line The 7th line The 8th line The 9th line The 10th line

1           1 1        2         1 1        3        3         1 1        4        6        4      1 1       5       10       10       5       1 1      6     15      20      15      6      1 1      7    21     35     35     21     7     1 1     8     28     56     70     56     28     8    1 1    9    36    84    126    126    84    36    9    1 1   10   45   120   210   252   210   120   45   10   1

Find the sum of following Pascal triangle numbers lines.

(vi)              The 8^th lines=2^7=128

(vii)             The 9^th lines=2^8=256

(viii)           The 10^th lines=2^9=512

(ix)              The 11^th lines=2^10=1024

(x)                The 12^th lines=2^11=2048