If the first differences of a sequence are a constant -7 and the third term is 22, Find the first 5 terms of the sequence

Question

anonymous · Answer

Here differnce d= -7
$$t_3= 22 \rightarrow n= 3$$
Let us find the first term a using the following formula.

$$t_n= a+(n-1)d \rightarrow t_3= a+(3-1)(-7) \rightarrow 22= a+2(-7)$$
$$\rightarrow 22= a-14 \rightarrow 22+14= a \rightarrow a= 36$$
Hence first term of the givemn sequence is a = 36

since 
$$t_n= a+(n-1)d \rightarrow t_n= 36+(n-1)(-7) \rightarrow   t_n= 36-7n+7 $$
$$\rightarrow   t_n= 43-7n$$
Now, in order to find first five terms of the given sequence, let us plu n= 1, 2, 3, 4, 5. in  $$t_n= 43-7n$$
so $$t_1= 43-7 \times 1= 43-7 = 36$$
$$t_2= 43-7 \times 2= 43-14 = 29$$
$$t_3= 43-7 \times 3= 43-21 = 22$$
$$t_4= 43-7 \times 4= 43-28 = 15$$
$$t_5= 43-7 \times5= 43-35 = 8$$
Hence first five terms of the given sequence are 36, 29, 22, 15 & 8.

@djinaocean