series help (1 Viewer)

[poopoohead]

Could some one please help me with the following questions please???

Q1:

The 20th term of arithmetic series is 131 and the sum of the 6th to 10th terms inclusive is 235. Find the sum of the first 20 terms

Q2: (similar sort of question)

The sum of the first 4 terms of an arithmetic series is 42 and the sum of the 3rd and 7th term is 46. Find the sum of the first 20 terms


Answers (from back of book): Q1: 1290
Q2: 1010

thankyou

 

[SoulSearcher]

Q1:
a + 19d = 131 ... (1)
10a + 190d = 1310 ... (2)
S₁₀ - S₅ = 235
5(2a+9d) - 5(2a+4d)/2 = 235
10(2a+9d) - 5(2a+4d) = 470 ... (3)
10a + 70d = 470 ... (4)
(2) - (4)
120d = 840
d = 7
Therefore, substituting d=7 into (1),
a + 133 = 131
a = -2
Therefore sum of first 20 terms is
S₂₀ = 20/2 * (-2+131)
= 10 * 129
= 1290

Q2:
2(2a+3d) = 42
4a + 6d = 42 ... (1)
(a+2d) + (a+6d) = 46
2a + 8d = 46 ... (2)
4a + 16d = 92 ... (3)
(3) - (1)
10d = 50
d = 5
Therefore, substituting d=5 into (1),
4a + 30 = 42
4a = 12
a = 3
Therefore sum of first 20 terms is
S₂₀ = 20/2 * (2*3 + 19*5)
= 10 * 101
= 1010

 

[poopoohead]

thankyou

 

[Jaguar33]

Q2:
2(2a+3d) = 42
4a + 6d = 42 ... (1)
(a+2d) + (a+6d) = 46
2a + 8d = 46 ... (2)
4a + 16d = 92 ... (3) (can anyone explain how he got to this line (like i know its multiplied by 2? but why does he do this to get to this third equation?)
(3) - (1)
10d = 50
d = 5
Therefore, substituting d=5 into (1),
4a + 30 = 42
4a = 12
a = 3
Therefore sum of first 20 terms is
S₂₀ = 20/2 * (2*3 + 19*5)
= 10 * 101
= 1010[/QUOTE]

 

[wandering17]

oh nvm i understand now.. lol

 

[keepLooking]

Simultaneous equation, so he can subtract a from the equation resulting in only one variable.