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 \(\quad r=-2\)