Question

if the fifth term of AP is 16 and the ninth term 28 then the 12 term is?​

Answer 1

Step-by-step explanation:

a5=a+4d=16

a9=a+8d=28

Solve both equations we get d=3,a=4

So we find 12th term of this AP

a12=a+11d=4+33=37

Answer 2

Given,

The fifth term of an AP$\[ = 16\]$

The ninth term of an AP$\[ = 28\]$

To find,

The twelfth term of the AP.  

Solution,

Assume that the first term of the AP is a and the common difference is d.

Know that the nth term is given as $\[{a_n} = a + \left( {n - 1} \right)d.\]$

Therefore,

$\[\begin{array}{l} \Rightarrow {a_5} = a + \left( {5 - 1} \right)d\\ \Rightarrow 16 = a + 4d - - - - - \left( 1 \right)\end{array}\]$

Similarly,

$\[\begin{array}{l} \Rightarrow {a_9} = a + \left( {9 - 1} \right)d\\ \Rightarrow 28 = a + 8d - - - - - \left( 2 \right)\end{array}\]$

Solve equations (1) and (2).

$\[\begin{array}{l} \Rightarrow a = 4\\ \Rightarrow d = 3\end{array}\]$

So, the twelfth term of the AP is

$\[\begin{array}{l} \Rightarrow 4 + \left( {12 - 1} \right)3\\ \Rightarrow 4 + \left( {11 \times 3} \right)\\ \Rightarrow 4 + 33\\ \Rightarrow 37\end{array}\]$

Hence, the value of the twelfth term of the AP is $\[37.\]$