maths

1.
T(1) = a
Common difference = r

T(4) :
a + 3d = 1 ...... [1]

T(11) :
a + 10d = -13 ...... [2]

[2] - [1] :
7d = -14
d = -2

T(18)
= a + 17d
= (a + 10d) + 7d
= -13 + 7(-2)
= -27


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2.
T(1) + T(2) + ...... + T(12)
= (12)² - 3(12)
= 108

T(1) + T(2) + ...... + T(11)
= (11)² - 3(11)
= 88

T(12)
= 108 - 88
= 20


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3.
X1 = a
Common difference = d

X2 :
a + d = 2 ...... [1]

X2 :
a + 5d = 14 ...... [2]

Xn :
a + (n - 1)d = 77 ...... [3]

[2] - [1] :
4d = 12
d = 3

Put d = 3 into [1] :
a + 3 = 2
a = -1

Put a = -1and d = 3 into [3] :
-1 + (n - 1) ´ 3 = 77
n = 27

Sum of the first n terms
= n(X1 + Xn)/2
= 27 ´ (-1 + 77) /2
= 1026


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4.
T(1), a = 162
Common ratio, r = 108/162 = 2/3

The nth term :
162 ´ (2/3)^(n - 1) = 1024/243
(2/3)^(n - 1) = 512/19683
(2/3)^(n - 1) = (2/3)^9
n - 1 = 9
n = 10
The number of terms = 10


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5.
X1 = a
Common ratio = r

Xn+3/Xn :
ar^(n+2) / ar^(n-1) = 8
r^3 = 8
r = 2

X4 + X5 + X6 :
a(2)^3 + a(2)^4 + a(2)^5 = 280
8a + 16a + 32a = 280
a = 5
Hence, X1 = 5


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6.
1/4 + 1/16 + 1/64 + ...... is the sum to infinity of a G.S.
T(1), a = 1/4
Common ratio, r = (1/16) / (1/4) = 1/4

5/8 + 5/32 + 5/128 + ...... is the sum to infinity of a G.S.
T(1), a = 5/8
Common ratio r = (5/32) / (5/8) = 1/4

1/4 - 5/8 + 1/16 - 5/32 + 1/64 - 5/128 + ...
= (1/4 + 1/16 + 1/64 + ......) - (5/8 + 5/32 +5/128 + ......)
= (1/4) / [1 - (1/4)] - (5/8) / [1 - (1/4)]
= 1/3 - 5/6
= -1/2


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7.
T(1) = a
Common ratio = r

The 4th term:
ar^3 = 64/3 ...... [1]

The 7th term:
ar^6 = 512/81 ...... [2]

[2]/[1] :
r^3 = 8/27
r = 2/3

Put r = 2/3 into [1] :
a (2/3)^3 = 64/3
a = 72

Sum to infinity of the sequence
= a/(1 - r)
= 72 / [1 - (2/3)]
= 215