AP H2 Maths

Destined2REIGN SG

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.

unclebutcher SG

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

unclebutcher SG

C.) to prove it’s an AP, d must be constant. that’s how you prove. but forgot already.

Chemfreak022 SG

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.

Destined2REIGN SG

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?

Chemfreak022 SG

(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

wee_ws SG

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