Question

The 5th term of AP is 11 and 9 th term is 7 find the 16 the term​

Answer 1

Answer:

a+4d=11

a+8d=7

using elimination method

a + 8d =7

a + 4d =11

- - -

4d =-4

d =-1

a +(-4)=11

a=15

a+15 d

=0

Answer 2

Answer:

$\underline{\boxed{{a}_{16} = 0}}$

Step-by-step explanation:

$\underline{\underline{\textbf{Given that...}}}$

The 5th term of AP is 11 and 9 th term is 7. Now we need to find 16th term

$\underline{\underline{\textbf{Solution...}}}$

Now let,

First term = a

and

Common Difference = d

Now from Given we have...

$\mathsf{{a}_{5} \:= 11}$ ----------- (1)

$\mathsf{{a}_{9} \:= 7}$ ------------ (2)

Now we know that...

$\mathsf{{a}_{5} \:= a + 4d}$

and

$\mathsf{{a}_{9} \:= a + 8d}$

From Equation 1 and 2 we have..

$\mathsf{a + 4d = 11}$

$\implies \mathsf{a = 11-4d}$ ---------- (3)

$\mathsf{a + 8d = 7}$ ------------- (4)

Now putting value of a = 11-4d in eq 4 we have

$\mathsf{11 -4d + 8d = 7}$

$\implies \mathsf{11 + 4d = 7}$

$\implies \mathsf{4d = -4}$

$\implies \mathsf{d = (-1) }$

Therefore now putting value of d = (-1) in eq 3 we have

$\mathsf{a = 11-4(-1)}$

$\implies \mathsf{a = 11 + 4 }$

$\implies \mathsf{a = 15}$

Now to find 16th term we have formula

$\boxed{\mathsf{\bigstar \: a_n = a + (n-1)d \: \bigstar }}$

Where,

$\mathsf{a_n =\: nth\: term}$

$\mathsf{a =\: first \: term\: = 15 }$

$\mathsf{n = \: Number\: of\: terms = 16}$

and

$\mathsf{d = \: Common \: divisor = -1 }$

Therefore we have

$\mathsf{{a}_{16} = 15 + (16-1)(-1)}$

$\implies \mathsf{{a}_{16} = 15 + (15)( -1) }$

$\implies \mathsf{{a}_{16} = 15 -15 }$

$\implies \mathsf{{a}_{16} = 0 }$

$\underline{\textbf{Therefore required answer is..}}$

$\boxed{\boxed{\mathsf{{a}_{16} = 0}}}$