## Question 10(a)

### Steps of Solution

\( a + 4d = 11 \)

\( a + 6d = 3(a + d) \)

\( a + 6d = 3a + 3d \)

\( a + 6d – 3a – 3d = 0 \)

\( -2a + 3d = 0 \)

\( 3d = 2a \)

\( a = \frac{3d}{2} \)

\( \frac{3d}{2} + 4d = 11 \)

\( \frac{3d + 8d}{2} = 11 \)

\( \frac{11d}{2} = 11 \)

\( 11d = 22 \)

\( d = 2 \)

\( a = \frac{3 \times 2}{2} \)

\( a = 3 \)

### Explanation of the Steps

Given the 5th term of the arithmetic progression is 11, we use the formula for the \( n \)-th term of an arithmetic progression, \( a_n = a + (n-1)d \). Here, \( a \) is the first term and \( d \) is the common difference. Therefore, the 5th term equation is:

\( a + 4d = 11 \)

We also know that the 7th term is three times the 2nd term. Using the \( n \)-th term formula for the 7th and 2nd terms, we get:

7th term: \( a + 6d \)

2nd term: \( a + d \)

The equation provided is:

\( a + 6d = 3(a + d) \)

Expanding and simplifying the equation:

\( a + 6d = 3a + 3d \)

\( a + 6d – 3a – 3d = 0 \)

\( -2a + 3d = 0 \)

Solving for \( a \):

\( 3d = 2a \)

\( a = \frac{3d}{2} \)

Substituting \( a \) in the equation \( a + 4d = 11 \):

\( \frac{3d}{2} + 4d = 11 \)

Combining like terms:

\( \frac{3d + 8d}{2} = 11 \)

\( \frac{11d}{2} = 11 \)

Solving for \( d \):

\( 11d = 22 \)

\( d = 2 \)

Now, substituting \( d \) back into \( a = \frac{3d}{2} \):

\( a = \frac{3 \times 2}{2} \)

\( a = 3 \)

Thus, the first term \( a \) is 3 and the common difference \( d \) is 2.

## Question 10(b)

### Steps of Solution

A.P.:

\(

\begin{aligned}

a & =3 \\

T_2 & =3+(2-1) d \\

& =3+d \\

T_6 & =3+(6-1) d \\

& =3+5 d

\end{aligned}

\)

G.P.:

\(

\begin{aligned}

a & =3 \\

T_3 & =(3) r^{3-1} \\

& =3 r^2 \\

T_5 & =(3) r^{5-1} \\

& =3 r^4 \\

3+d & =3 r^2 \\

r^2 & =\frac{3+d}{3}

\end{aligned}

\)

\(

\begin{aligned}

& 3+5 d=3 r^4 =3\left(r^2\right)^2 \\

& 3+5 d=3\left(\frac{3+d}{3}\right)^2 \\

& 3+5 d=\frac{3\left(9+6 d+d^2\right)}{9} \\

& 3(3+5 d)=9+6 d+d^2 \\

& d^2+6 d-15 d+9-9=0 \\

& d^2-9 d=0 \\

& d(d-9)=0 \\

& d=9 \\

&

\end{aligned}

\)

Ignore \(d=0\)

\(

\begin{aligned}

r^2 & =\frac{3+(9)}{3} \\

r^2 & =4 \\

r & =2

\end{aligned}

\)

### Explanation of the Steps

Arithmetic Progression (A.P.):

1. First term of the A.P.:

\( a = 3 \)

2. Second term of the A.P. (\( T_2 \)):

\( T_2 = 3 + (2-1)d \)

\( T_2 = 3 + d \)

3. Sixth term of the A.P. (\( T_6 \)):

\( T_6 = 3 + (6-1)d \)

\( T_6 = 3 + 5d \)

Geometric Progression (G.P.):

1. First term of the G.P.:

\( a = 3 \)

2. Third term of the G.P. (\( T_3 \)):

\( T_3 = 3 \cdot r^{3-1} \)

\( T_3 = 3r^2 \)

3. Fifth term of the G.P. (\( T_5 \)):

\( T_5 = 3 \cdot r^{5-1} \)

\( T_5 = 3r^4 \)

4. Equating the second term of the A.P. and the third term of the G.P.:

\( 3 + d = 3r^2 \)

\( r^2 = \frac{3 + d}{3} \)

Solving for the Common Difference (\( d \)) and Common Ratio (\( r \)):

1. Equating the sixth term of the A.P. and the fifth term of the G.P.:

\( 3 + 5d = 3r^4 \)

\( 3 + 5d = 3(r^2)^2 \)

2. Substituting \( r^2 \) from earlier:

\( 3 + 5d = 3 \left( \frac{3 + d}{3} \right)^2 \)

\( 3 + 5d = 3 \cdot \frac{(3 + d)^2}{9} \)

\( 3 + 5d = \frac{3(9 + 6d + d^2)}{9} \)

3. Simplifying the equation:

\( 3(3 + 5d) = 9 + 6d + d^2 \)

\( 9 + 15d = 9 + 6d + d^2 \)

\( d^2 + 6d – 15d + 9 – 9 = 0 \)

\( d^2 – 9d = 0 \)

4. Factoring the quadratic equation:

\( d(d – 9) = 0 \)

\( d = 9 \)

(Ignore \( d = 0 \) as it is not valid in this context)

5. Substituting \( d = 9 \) back to find \( r \):

\( r^2 = \frac{3 + 9}{3} \)

\( r^2 = \frac{12}{3} \)

\( r^2 = 4 \)

\( r = 2 \)

(Ignore \( r = -2 \) as the common ratio must be greater than 1)

Therefore, the common difference \( d \) of the A.P. is 9, and the common ratio \( r \) of the G.P. is 2.

ignore \(\quad r=-2\)