Question

In a sequence of 21 terms, the first 11 terms are in A.P with common difference 2 and the last 11 terms are in G.P. with a common ratio of 2. If the middle term of the A.P. is equal to the middle term of G.P. Then find the middle term of the entire sequence.

Answer 1

Hint: Firstly we will find the middle term of A.P as well as G.P of the sequence. Then since it is given that the middle terms are equal we will equate them. Also the last term of A.P is equal to the first term of G.P therefore we will equate them and then solve the equations so formed to get the desired answer.
The nth term of A.P is given by:
\[{T_n} = a + \left( {n - 1} \right)d\]
Where a is the first term of AP and d is the common difference
 The nth term of G.P is given by:-
\[{T_n} = a{r^{n - 1}}\]
Where a is the first term of GP and r is the common ratio.

Complete step-by-step answer:
The middle term of a sequence having n terms is given by:-
\[{\text{Middle Term}} = \left\{ {\dfrac{{n + 1}}{2},{\text{n is odd}}} \right\}\]
\[{\text{Middle Term}} = \left\{ {\dfrac{n}{2},{\text{n is even}}} \right\}\]
Since the total number of terms in sequence are 21 and since 21 is odd
Then the middle term is given by:
\[{\text{Middle Term}} = \dfrac{{21 + 1}}{2}\]
\[ \Rightarrow {\text{Middle Term}} = \dfrac{{22}}{2}\]
\[ \Rightarrow {\text{Middle Term}} = 11\]
Hence 11th term of the sequence is its middle term.
Now we will find the middle term of AP
Since the total number of terms in AP is 11 therefore its middle term is given by:-
\[{\text{Middle Term}} = \dfrac{{11 + 1}}{2}\]
\[ \Rightarrow {\text{Middle Term}} = \dfrac{{12}}{2}\]
\[ \Rightarrow {\text{Middle Term}} = 6\]
Hence 6th term is the middle term of AP
Now we will find the 6th term of AP :
Since the nth term of an AP is given by:-
\[{A_n} = a + \left( {n - 1} \right)d\]
Also the common difference given is 2.
Hence, putting in the values we get:-
\[{a_{6{\text{ }}}} = a + (6 - 1)d\]
Simplifying it further and putting the value of d we get:-
\[{a_{6}}{\text{ }} = a + 5 \times 2\]
\[ \Rightarrow {a_6}_{{\text{ }}} = a + 10.....................\left( 1 \right)\]
Now, we will find the middle term of GP
Since the total number of terms in GP is 11 therefore its middle term is given by:-
\[{\text{Middle Term}} = \dfrac{{11 + 1}}{2}\]
\[{\text{Middle Term}} = \dfrac{{12}}{2}\]
\[ \Rightarrow {\text{Middle Term}} = 6\]
Hence 6th term is the middle term of GP
Now we will find the 6th term of GP :
Since the nth term of a GP is given by:-
\[{P_n} = a{r^{n - 1}}\]
Also the common ratio given is 2.
Hence,
\[{p_{6}}{\text{ }} = p{r^{6 - 1}}\]
Putting the value of r we get:-
\[{p_6}_{\text{ }} = p \times {2^5}.................\left( 2 \right)\]
Now since middle terms of AP and GP are equal
Therefore, equating equations 1 and 2 we get:
\[{a_6} = {p_{{\text{6}}}}\]
\[ \Rightarrow (a + 10) = p \times {2^5}............(3)\]
Also, \[11th\;term{\text{ }}of{\text{ }}A.P\; = \;1st\;term{\text{ }}of{\text{ }}G.P\]
Therefore,
\[{a_{11}} = {p_1}\]
\[ \Rightarrow a + (11 - 1) \times 2 = p \times {r^{1 - 1}}\]
\[ \Rightarrow a + (10) \times 2 = p \times 1\]
\[ \Rightarrow a + 20 = p...................(4)\]
Now solving equations 3 and 4 we get:-
Putting the value of p from equation 4 in equation 3 we get:
\[a + 10 = \left( {a + 20} \right) \times {2^5}\]
Solving it further we get:-
\[a + 10 = \left( {a + 20} \right) \times 32\]
Simplifying it further we get:-
\[a + 10 = 32a + 640\]
\[32a - a = 10 - 640\]
\[ \Rightarrow 31a = - 630\]
\[ \Rightarrow a = \dfrac{{ - 630}}{{31}}\]
Putting the value of a in equation 4 we get:-
\[\dfrac{{ - 630}}{{31}} + 20 = p\]
\[ \Rightarrow \dfrac{{ - 630 + 620}}{{31}} = p\]
\[ \Rightarrow \dfrac{{ - 10}}{{31}} = p\]
Therefore middle term of the sequence is 11th of A.P or 1st term of G.P i.e, \[p = \dfrac{{ - 10}}{{31}}\]

Note: Arithmetic sequence is a sequence in which the difference between each of the consecutive terms is equal.
Geometric sequence is a sequence in which the ratio of a pair of consecutive numbers is the same as that of another pair.
The middle term of a sequence having n terms is given by:-
\[{\text{Middle Term}} = \left\{ {\dfrac{{n + 1}}{2},{\text{n is odd}}} \right\}\]
\[{\text{Middle Term}} = \left\{ {\dfrac{n}{2},{\text{n is even}}} \right\}\]