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The progression is : 2,5,8,11,14,17

The 7th term of an AP is four times its 2nd term

12th term is two more than three times of its 4 th term.

The progression.

First term of an AP =a

Common difference=d

Tn = n th term

a = First term

n = Number of terms

d = Common difference

The 7th term of an AP is Four times its 2nd term.

⟹T₇ = 4 T₂

⟹a + (7 - 1) d = 4 { a +(2 - 1)d}

⟹a + 6d = 4 {a + d}

⟹4a - a = 6d - 4d

⟹3a = 2d_________(i)

12th term is 2 more than three times of its 4 th term.

⟹T₁₂=3T₄+2

⟹a + (12 - 1) d = 3{a +(4 - 1)d}+2

⟹a + 11d = 3{a + 3d} + 2

⟹a + 11d = 3a + 9d + 2

⟹11d - 9d = 3a - a + 2

⟹2d = 2a + 2

⠀⠀⠀[From eqⁿ (i) 2d = 3a]

⟹3a = 2a + 2

⟹3a - 2a = 2

First term = 2

In equation (i)

⟹2d = 3a

⟹2d = 3×2

⠀⠀⠀⠀⠀[The value of a=2]

⟹2d = 6

⟹d = 3

Common difference = 3

⟹T₁ = a = 2

⟹T₂ = a + d = 2 + 3 = 5

⟹T₃ = a + 2d = 2 + 6 = 8

⟹T₄ = a + 3d = 2 + 9 = 11

⟹T₅ = a + 4d = 2 + 12 = 14

⟹T₆ = a + 5d = 2 + 15 = 17

i think the answer is seven, for if i'm being useless °ω°

step-by-step explanation:

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!