Question

If 3rd and 9th term of ap are 4 and -8. Find the 12 th term

Answer 1

Given :-

3rd term of the AP is 4

9th term of the AP is - 8

Required to find :-

• 12th term of the AP

Formula used :-

$\Large{\dagger{\boxed{\rm{ {a}_{nth} = a + ( n - 1 ) d }}}}$

Solution :-

Given information :-

3rd term of the AP is 4

9th term of the AP is - 8

We need to find the 12th term of the AP

So,

Let's consider the given information ;

3rd term = 4

But,

3rd term can be represented as ' a + 2d '

So,

• a + 2d = 4 $\orange\rightarrowtail{\text{Equation - 1 }}$

Consider this as equation - 1

Similarly ,

9th term = - 8

But,

9th term can be represented as ' a + 8d '

So,

• a + 8d = - 8 $\orange\rightarrowtail{\text{Equation - 2 }}$

Consider this as equation - 2

Now,

Let's solve these 2 equations using elimination method

Hence,

Subtract equation 1 from equation 2

$\tt a + 8d = - 8 \\ \tt a + 2d = \: \: \: \: 4 \\ \rm \underline{(-)( - ) \: \: \: \: ( - ) \: \: } \\ \tt \underline{ \: \: \: \: \: + 6d = - 12}$

This implies ;

=> 6d = - 12

=> d = -12/6

=> d = - 2

So,

• Common difference ( d ) = - 2

Substitute the value of d in equation 1

a + 2d = 4

a + 2 ( - 2 ) = 4

a - 4 = 4

a = 4 + 4

a = 8

So,

• First term ( a ) = 8

Using the formula ;

$\Large{\dagger{\boxed{\rm{ {a}_{nth} = a + ( n - 1 ) d }}}}$

$\rightarrow{\tt{ {a}_{nth} = {a}_{12} }}$

$\rightarrow{\tt{ {a}_{12} = 8 + ( 12 - 1 ) - 2 }}$

$\rightarrow{\tt{ {a}_{12} = 8 + ( 11 ) - 2 }}$

$\rightarrow{\tt{ {a}_{12} = 8 + ( - 22 ) }}$

$\rightarrow{\tt{ {a}_{12} = 8 - 22 }}$

$\rightarrow{\tt{ {a}_{12} = - 14 }}$

Therefore ,

12th term = - 14

Answer 2

Let ,

The first term and common difference of AP be " a " and " d "

Given ,

The third term of AP is 4 i.e

a + 2d = 4 --- (i)

And the ninth term of AP is -8 i.e

a + 8d = -8 --- (ii)

Subtract eq (i) from eq (ii) , we get

6d = -12

d = -2

Put the value of d = -2 in eq (i) , we get

a + 2(-2) = 4

a - 4 = 4

a = 8

Now , the 12th term of AP will be

$\sf \mapsto a_{12} = 8 + 11( - 2) \\ \\ \sf \mapsto a_{12} = - 14$

$\sf \therefore \underline{The \: 12th \: term \: of \: AP \: will \: be \: -14}$