AP H2 Maths
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A)Some qn that i need help in doing.
if y is a positive interger, show that the sum of the series
y + (y+3) + (y+6) + ... + 4y is 5 times the sum of the series
1+2+3+...+y
Still dun get this.
Got one more. An AP has n terms. First term is 2 and common difference is 2/3. If sum of last four terms is 72 more than the sum of the first for terms, find n. Ans is 31. -
B.) a+ (a+d) + (a+2d) + .....+ (a+ (n-1)d)= 3n^2 sum of ap: (n/2) (2a + (n-1)d) I THINK. FORGOT ALREADY
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C.) to prove it's an AP, d must be constant. that's how you prove. but forgot already. to find the sum.....use the sum of ap formula a= log a d = r(if im not wrong, forgot the laws of log already zzz) ZZZ. after attempting this qn, i realised i forgot alot already. sigh. only been 3 months. HAHA
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1+ 2 + 3+ 4 +...+ y = y(y+1)/2
y + (y+3) + (y+6) + (y+3y) = y + (y+y+ y + y +.....y) + (3 + 6 + 9 + ... + 3y)
= y^2 +y + 3( 1+ 2 + 3 + ...+y) = y^2 + y + 3y(y+1)/2
= 2y^2/2+3y^2/2+2y/2 +3y/2
= 5y(y+1)/2.
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Originally posted by Chemfreak022:
1+ 2 + 3+ 4 +...+ y = y(y+1)/2
y + (y+3) + (y+6) + (y+3y) = y + (y+y+ y + y +.....y) + (3 + 6 + 9 + ... + 3y)
= y^2 +y + 3( 1+ 2 + 3 + ...+y) = y^2 + y + 3y(y+1)/2
= 2y^2/2+3y^2/2+2y/2 +3y/2
= 5y(y+1)/2.
how did you get this?
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(y+3) + (y+6) + ...+ (y+3y) ---------> y terms here
The 3+6...+3y is taken out, leaving y terms of y = y x y = y^2
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Hi,
For (A), LHS is an AP with first term y and common difference 3. You need to find the number of terms, to find the sum. RHS is an AP with first term 1 and common difference 1 and it has y terms.
For (B), you need to use the fact that
T_n = S_n - S_(n - 1) and
show that
T_n / T_(n - 1) is a constant.
For (C), try to recognise the general term T_n from the given sum and show that
T_n - T_(n - 1) is a constant. Finding the sum is easy.
Keep trying!
Cheers,
Wen Shih