(1) let a[ 1]=a, and the tolerance d(d≠0) is known.

A+(a+d)(a+2d)=2 1, that is, a+d=7 (1).

(a+5d)？ =a(a+20d) (2)

A = 5 and d = 2 are obtained from (1)(2).

So a[n]=2n+3.

(2) let c[n]= 1/b[n]

Then c[ 1]=3.

And when n≥2, c [n]-c [n-1] = a [n-1] = 2n+1.

That is, c[n]-c[n- 1]=2n+ 1.

c[n]=(c[n]-c[n- 1])+(c[n- 1]-c[n-2])+...+(c[2]-c[ 1])+c[ 1]

=(2n+ 1)+(2(n- 1)+ 1)+...+(2 2+ 1)+3

=n(3+(2n+ 1))/2

=n？ +2n

b[n]= 1/(n？ +2n)=( 1/2)(( 1/n)-( 1/(n+2)))

t[n]=( 1/2)(( 1-( 1/3))+(( 1/2)-( 1/4)+...+(( 1/(n- 1))-( 1/(n+ 1))+(( 1/n)-( 1/(n+2)))

=( 1/2)( 1+( 1/2)-( 1/(n+ 1))-( 1/(n+2)))

=(3/4)-(2n+3)/(2n？ +6n+4)

So T[n]=(3/4)-(2n+3)/(2n? +6n+4)

I hope I can help you!