Question

In a sequence of 9 terms, the first 5 terms are in A.P whose common difference is 2 and the last 5 terms are in GP whose common ratio is 1/2. If the middle terms of the AP and GP are equal, then the middle term of the GP is
A) 1/3 B) 4/3 C) 5/3 D) 7/3​

Answer 1

Answer:

4/3

Step-by-step explanation:

d = 2, r = 1/2

Consider the sequence as $a,a+d,a+2d,a+3d,(a+4d)r,(a+4d)r^2,(a+4d)r^3,(a+4d)r^4$

According to the question, midterm of AP = midterm of G.P

$a+2d=(a+4d)r^2\\a+2(2)=(a+4(2)(\frac{1}{2})^2)\\4a + 16 = a+ 18\\3a = -8, a = \frac{-8}{3}$

Midterm

= a + 2d

= -8/3 + 2

= 4/3

Answer 2

Given: In a sequence of 9 terms, the first 5 terms are in A.P whose common difference is 2 and the last 5 terms are in GP whose common ratio is $\frac{1}{2}$. the middle terms of the AP and GP are equal.

To find: the middle term of the GP.

Solution:

Know that from the question, $\[d = 2,r = \frac{1}{2}\]$.

Understand that the sequence will be as follows:

$\[a,a + d,a + 2d,a + 3d,\left( {a + 4d} \right)r,\left( {a + 4d} \right){r^2},\left( {a + 4d} \right){r^3},\left( {a + 4d} \right){r^4}\]$

Observe that from the question, middle terms of the AP and GP are equal.

Therefore, find the value of a.

$\[\begin{array}{l}a + 2d = \left( {a + 4d} \right){r^2}\\ \Rightarrow a + 2\left( 2 \right) = \left( {a + 4\left( 2 \right)} \right) \times \frac{1}{2}\\ \Rightarrow 4a + 16 = a + 18\\ \Rightarrow 3a = - 8\end{array}\]$

$\[ \Rightarrow a = \frac{{ - 8}}{3}\]$

FInd the middle term of the GP.

Middle term$\[ = a + 2d\]$

                   $\[ = - \frac{8}{3} + 2 \times 2\]$

                   $\[ = \frac{{ - 8 + 12}}{3}\]$

                   $\[ = \frac{4}{3}\]$

Hence, the middle term of the GP is $\frac{4}{3}$.