Chapter 11.3, Problem 1ST

a.

To determine

To find: The missing terms of the sequence.

Answer to Problem 1ST

The given sequence is a Geometric progression (G.P) and the fifth and sixth terms are $\frac{27}{8}\text{ and }\frac{81}{16}$ respectively

Explanation of Solution

Given information: $\frac{2}{3},1,\frac{3}{2},\frac{9}{4},\frac{?}{},\frac{?}{}$

From the above sequence the common ratio of the two terms are constant. Therefore, the given sequence is a Geometric progression (G.P).

Calculation:

Let, first term, $a_1$ = $\frac{2}{3}$

Second term, $a_2$ = $1$

Third term, $a_3$ = $\frac{3}{2}$

As we know that, the common ratio, $r=\frac{a_2}{a_1}=\frac{a_3}{a_2}$

Now,

  $\begin{array}{l}\Rightarrow r=\frac{1}{\frac{2}{3}}=\frac{\frac{3}{2}}{1}\\ \Rightarrow r=\frac{3}{2}=\frac{3}{2}\end{array}$

Hence, the given sequence is a Geometric progression(G.P).

As we know that, general series of G.P: $a_1,a_{1}r,a_1{r}^2,\mathrm{...}$

Therefore, fifth term,

  $\begin{array}{l}\Rightarrow a_5=a_1{r}^4\\ \Rightarrow a_5=\frac{2}{3}\times {\left(\frac{3}{2}\right)}^4\\ \Rightarrow a_5=\frac{27}{8}\end{array}$

Therefore, sixth term,

  $\begin{array}{l}\Rightarrow a_6=a_1{r}^5\\ \Rightarrow a_6=\frac{2}{3}\times {\left(\frac{3}{2}\right)}^5\\ \Rightarrow a_6=\frac{81}{16}\end{array}$

Hence, the fifth and sixth terms are $\frac{27}{8}\text{ and }\frac{81}{16}$ respectively.

b.

To determine

To find: The missing terms of the sequence.

Answer to Problem 1ST

The given sequence is an Arithmetic Progression (A.P).and the fifth and sixth terms are $\left(-11\right)\text{ and }\left(-15\right)$ respectively.

Explanation of Solution

Given information: $5,1,-3,-7,\frac{?}{},\frac{?}{}$

From the above sequence the common difference between the two terms are constant. Therefore, the given sequence is an Arithmetic Progression (A.P). Calculation:

Let, first term, $a_1$ = 5

Second term, $a_2$ = 1

Third term, $a_3$ = -3

As we know that, the common difference, $d=a_2-a_1=a_3-a_2$

Now,

  $\begin{array}{l}\Rightarrow d=\left(1\right)-\left(5\right)=\left(-3\right)-\left(1\right)\\ \Rightarrow d=-4=-4\end{array}$

Hence, the given sequence is an Arithmetic Progression (A.P)

As we know that, the general series of A.P: $a_1,a_1+d,a_1+2d,a_1+3d,\mathrm{....}$

Therefore, fifth term,

  $\begin{array}{l}\Rightarrow a_5=a_1+4d\\ \Rightarrow a_5=5+4\times \left(-4\right)\\ \Rightarrow a_5=5-16\\ \Rightarrow a_5=-11\end{array}$

Therefore, sixth term,

  $\begin{array}{l}\Rightarrow a_6=a_1+4d\\ \Rightarrow a_6=5+5\times \left(-4\right)\\ \Rightarrow a_6=5-20\\ \Rightarrow a_6=-15\end{array}$

Hence, the fifth and sixth terms are $\left(-11\right)\text{ and }\left(-15\right)$ respectively.

c.

To determine

To find: The missing terms of the sequence.

Answer to Problem 1ST

There is neither Arithmetic Progression (A.P) nor Arithmetic Progression (G.P) in the given sequence. Since the missing terms cannot be calculated.

Explanation of Solution

Given information: $\frac{1}{3},\frac{2}{5},\frac{3}{7},\frac{4}{9},\frac{?}{},\frac{?}{}$

From the above sequence the common difference between the two terms are not same or the common ratio is also not same. The sequence neither Arithmetic Progression (A.P) nor Arithmetic Progression (G.P).

Calculation:

Let, first term, $a_1$ = $\frac{1}{3}$

Second term, $a_2$ = $\frac{2}{5}$

Third term, $a_3$ = $\frac{3}{7}$

As we know that, the common difference, $d=a_2-a_1=a_3-a_2$

Now,

  $\begin{array}{l}\Rightarrow d=\frac{2}{5}-\frac{1}{3}=\frac{3}{7}-\frac{2}{5}\\ \Rightarrow d=\frac{6-5}{15}=\frac{15-14}{35}\\ \Rightarrow d=\frac{1}{15}\ne \frac{1}{35}\end{array}$

Hence, the given sequence not Arithmetic Progression (A.P)

Now,

As we know that, the common ratio, $d=\frac{a_2}{a_1}=\frac{a_3}{a_2}$

Now,

  $\begin{array}{l}\Rightarrow d=\frac{\frac{2}{5}}{\frac{1}{3}}=\frac{\frac{3}{7}}{\frac{2}{5}}\\ \Rightarrow d=\frac{2}{5}\times \frac{3}{1}\ne \frac{3}{7}\times \frac{5}{2}\\ \Rightarrow d=\frac{6}{5}\ne \frac{15}{14}\end{array}$

Hence, the given sequence not Geometric Progression (G.P)

d.

To determine

To find: The missing terms of the sequence.

Answer to Problem 1ST

The given sequence is an Arithmetic Progression (A.P).and the fifth and sixth terms are $\text{2}\text{.2 and 2}\text{.5}$ respectively.

Explanation of Solution

Given information: $1,1.3,1.6,1.9,\frac{?}{},\frac{?}{}$

From the above sequence the common difference between the two terms are constant. Therefore, the given sequence is an Arithmetic Progression (A.P). Calculation:

Let, first term, $a_1$ = 1

Second term, $a_2$ = 1.3

Third term, $a_3$ = 1.6

As we know that, the common difference, $d=a_2-a_1=a_3-a_2$

Now,

  $\begin{array}{l}\Rightarrow d=\left(1.3\right)-\left(1\right)=\left(1.6\right)-\left(1.3\right)\\ \Rightarrow d=0.3=0.3\end{array}$

Hence, the given sequence is an Arithmetic Progression (A.P)

As we know that, the general series of A.P: $a_1,a_1+d,a_1+2d,a_1+3d,\mathrm{....}$

Therefore, fifth term,

  $\begin{array}{l}\Rightarrow a_5=a_1+4d\\ \Rightarrow a_5=1+4\times \left(0.3\right)\\ \Rightarrow a_5=1+1.2\\ \Rightarrow a_5=2.2\end{array}$

Therefore, sixth term,

  $\begin{array}{l}\Rightarrow a_6=a_1+4d\\ \Rightarrow a_6=1+5\times \left(0.3\right)\\ \Rightarrow a_6=1+1.5\\ \Rightarrow a_6=2.5\end{array}$

Hence, the fifth and sixth terms are $\text{2}\text{.2 and 2}\text{.5}$ respectively.