5 maths questions about AS and GS

1. Equation of general term of a geometric sequence: ar^(n-1)

a= first term ; r= ratio ; n= no. of term


T4 x T10 = 27
ar^(4-1) x ar^(10-1) = 27
ar^3 x ar^9 = 27
(a^2) (r^12) = 27
(ar^6)^2 = 27

where term 7 is ar^6

therefore, T7 = square root of 27
= 3 squareroot 3



2. T(n) = ar^n-1
T (2n) = ar^ 2n-1

where it's given that the first term is -1/2 and the ratio is -2.
therefore, the 2n term is (-1/2)(-2)^2n-1
as 2n-1 is always an odd no.
thus there will be a negative sign in front of 2.
which cancel with the -ve sign in front of (1/2)

= (2)^-1 (2)^2n-1 <--1/2 = 2^-1
= 2^2n





3. Isn't it the last term : 1/a^(3n +6) ?

4. T(8) = a + (n-1) d = a + 7d = 11
a = 11- 7d ------*

Sum (15) = n/2 (2a+ (n-a)d )= 15/2 (2a + 14d )
sub * into this
Sum (15) = 15/2 (22- 14d + 14d)
= 165



5. Sum of the 1st GS fro 20 terms : S (n) = a (r^n - 1)/ (r-1)
S (20) = 2 (3^20-1)/ (2)
= 3^20 -1


As the 2nd sequence is +1 for each term in the 1st sequence, therefor there'll be + 20 for the sum of first 20 terms. Which is : 3^20-1 + 20 = 3^20 -19

功課應該自己做

1. Let T1=A*B, TN=A*B^N
T7=A*B^7
T4*T10=A*B^4*A*B^10=(A*B^7)^2=(T7)^2
Therefore, T7 equals to squareroot of 27.

2.問題不清

3.TN=a^(-3(N-1))
1/a^(3n+6)=a^(-3((n+3)-1))
Number of term=N=n+3

4.Let T1=A+B, TN=A+NB
T8=A+8B=11
summation T1 to T15=15(A)+((1+15)*15/2)B=15A+15(8B)=15(T8)=15(11)=165

5. 2+6+18+....+2*3^19=2(3^20-1)/(3-1)=3^20-1
3+7+19....+(2*3^19+1)=(3^20-1)+20=3^20+19