H2 Maths - AP/GP
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i neeed help (: haha
1. in an AP , 8th term is twice the 4th term and the 20th term is 40. Find the common diff an the sum of the terms from the 8th to the 20th term.
2. The rth term of an AP is (1+4r) Find in terms of n, the sum of the first n terms.
3. An arithmetic series has first term 1000 and common diff -1.4. Calculate the 1st negative term and the sum of all the positive terms
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1) Given: U8 = 2U4, U20 = 40
Apply nth term formula:
a + 7d = 2(a +6d) ........... a+5d =0 -(1)
a+19d =40 -(2)
solve simultaneously
.
.
.
2nd part of Q : S20-S7 (using the a and d you've found)
2)Ur= 1+4r
Let r=1, U1 =5
Let 3=2, U2=9
d=4
.....
3) Given: a =1000 d= -1.4
use formula
Sn = n/2[2000 +1.4(n-1)] <0
... find n
then should easy from here on... -
hey thanks alot!!!!:D i understand already!
this one quite weird qn, can help me explain the question. i don't understand the qn. btw is the 2002 a lvl paper 1 qn 10.
an ap in which the common diff is twice the 1st term is called a special ap Show that the sum of the first n terms of a special ap with 1st term a is n(square)a.
The (M-2)th term of a special ap is 663 and the Mth term is 731. Find M and find also the sum of the first M terms.
The sum of the first N terms of another special ap is S. FInd in the terms of S, the sum of the next N terms.
Answer: 22,8228, 35
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Originally posted by Occam's Razor:
1) Given: U8 = 2U4, U20 = 40
Apply nth term formula:
a + 7d = 2(a +6d) ........... a+5d =0 -(1)
a+19d =40 -(2)
solve simultaneously
.
.
.
2nd part of Q : S20-S7 (using the a and d you've found)
2)Ur= 1+4r
Let r=1, U1 =5
Let 3=2, U2=9
d=4
.....
3) Given: a =1000 d= -1.4
use formula
Sn = n/2[2000 +1.4(n-1)] <0
... find n
then should easy from here on...um can clarify fr the first question.
a + 7d = 2(a +6d) ........... a+5d =0 -(1)
shouldn't it be (a+3d) for a4?
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Originally posted by Occam's Razor:
1) Given: U8 = 2U4, U20 = 40
Apply nth term formula:
a + 7d = 2(a +6d) ........... a+5d =0 -(1)
a+19d =40 -(2)
solve simultaneously
.
.
.
2nd part of Q : S20-S7 (using the a and d you've found)
2)Ur= 1+4r
Let r=1, U1 =5
Let 3=2, U2=9
d=4
.....
3) Given: a =1000 d= -1.4
use formula
Sn = n/2[2000 +1.4(n-1)] <0
... find n
then should easy from here on...ummm. can you help me to do the first part about the negative number? i'm not very sure about that part.
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Originally posted by chocolates-xed:
um can clarify fr the first question.
a + 7d = 2(a +6d) ........... a+5d =0 -(1)
shouldn't it be (a+3d) for a4?
Yup you are right
Time to add questions to examworld
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Originally posted by chocolates-xed:
ummm. can you help me to do the first part about the negative number? i'm not very sure about that part.
Tn = a + (n-1)d
a = 1000, d = -1.4
Thus, when Tn < 0
1000 + (n-1)(-1.4) < 0
1.4n > 1001.4
n > 715.286Hence, the first negative n term is when n = 716
First negative n term = 1000 + (716 - 1)(-1.4) = -1
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Originally posted by chocolates-xed:
hey thanks alot!!!!:D i understand already!
this one quite weird qn, can help me explain the question. i don't understand the qn. btw is the 2002 a lvl paper 1 qn 10.
an ap in which the common diff is twice the 1st term is called a special ap Show that the sum of the first n terms of a special ap with 1st term a is n(square)a.
The (M-2)th term of a special ap is 663 and the Mth term is 731. Find M and find also the sum of the first M terms.
The sum of the first N terms of another special ap is S. FInd in the terms of S, the sum of the next N terms.
Answer: 22,8228, 35
Let first term = a, common difference = 2a
Sn = (n/2)*(2a + (n-1)(2a))
= (n/2)*(2a + 2an - 2a)
= n²a (shown)Mth term - (M-2)th term = 2 * common difference
731 - 663 = 2 * 2a
a = 17, d = 34
731 = 17 + (M-1)(34)
M = 22Sum of first M terms =22² * 17 = 8228
Sn = n²a
S2n = 4n²aDifference of next N terms = S2n - Sn
= 3n²a
= 3S ===> (not 35 hor) -
Originally posted by eagle:
Tn = a + (n-1)d
a = 1000, d = -1.4
Thus, when Tn < 0
1000 + (n-1)(-1.4) < 0
1.4n > 1001.4
n > 715.286Hence, the first negative n term is when n = 716
First negative n term = 1000 + (716 - 1)(-1.4) = -1
err how to find the sum of alll the positive terms?
the ans is 357643
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first negative term is when n = 716
so last positive term is when n = 715 lor
So sum = (715/2) (2000 + 714 * -1.4) {2a + (n-1)d}
you should get the answer