sequence (urgent)

3.
T7=sum7-sum6
=(7^2+3x7) - (6^2+3x6)
=16

24.
Sn=n^2
S1=T1=1

Sn =(n/2)(T1+Tn)=n^2
S10=(10/2)(1+Tn)=10^2
5(1+Tn)=100
Tn=19

27.
ab=100^2
b=10000/a

arithmetic mean of log a and log b:
(1/2)(log a + log b)
=(1/2)(log a + log 10000/a)
=(1/2)(log a + log 10000 - log a)
=(1/2)(4)=2

35
I. log a^2, log b^2 , log^c2
=>2log a, log ac , 2log c
=>2log a, (log a+log c) , 2log c
=>log a, (log a + log c)/2 , log c
correct

II. a^3, b^3, c^3
=>a^3, √(a c)^3 , c^3
=>a^3, √(a^3 x c^3) , c^3
correct

III. 4^a, 4^b, 4^c
=>4^a, 4^(√(a c), 4^c
If it is a geometric sequence,4^b should be √[(4^a)(4^c)]=4^[(a+c)/2]
wrong

40.
I.difference of T1T2,T2T3,T3T4 are all (a+3)
correct

II.When b=1,
7,62,509,4092
difference of T1T2,T2T3,T3T4 are different.
wrong

III.
log c^3, log c^8, log c^13, log c^18
=>3log c, 8log c,13log c, 18log c
difference of T1T2,T2T3,T3T4 are all 5log c
correct

3. a_n=n^2+3n-(n-1)^2-3(n-1)
=2n+2
a_7=2*7+2=16

8. are in GP, so q^2=pr
(i) (kq)^2=k^2q^2=k^2pr
(kp)(kr)=k^2pr, so (kq)^2=(kp)(kr)
and (kp), (kq), (kr), (ks) are in GP
(ii) (a^q)^2=a^2q
a^(p)*a^(q)=a^(p+q), and 2q≠p+q
so it's not in GP
(iii) loq(q)=log√ (pr)=1/2log(pr)=1/2(log(p)+log(r))
so it is in AP

24. a_n=n^2-(n-1)^2=2n-1
a_10=2*10-1=19

27. ab=10000
(loga+logb)/2=(logab)/2=(log10000)/2=4/2=2

35.
(i)loga^2, logac, log^c2 ==> 2loga, loga+logc, 2logc
2(loga+logc)=2loga+2logc
(ii)b^3=√ (ac)^3=(ac)^(3/2)
√ (a^3)(c^3)=(ac)^(3/2), so it is.
(iii)(4^b)^2=4^2b
(4^a)(4^c)=4^(a+c), a+c≠2b, so it's not

40.
(i)公差=(a-3)
(ii) b=1代入發現不合
(iii) 公差=5logc

3. D
solution:n=7
7^2+3(7)
=49+21
=70

8.E
n=10
10^2
=100

i can only do these two...sorry