OCR A level C2 maths, Jan06 Q5

5) In a geometric progression, the first term is 5 and the second term is 4.8

(ii) The sum of the first n terms is greater than 124, show that:
0.96(to the power of n)<0.008

and use logaritms to calculate the smallest possible value of n.

6 mark qu and I have no idea how to do it, I like this topic but this question and the way its worded is confusing

Reply 1

FoOtYdUdE

ok, this is easy so dw mate.

sn = [ a(1-r^n) ] / [ 1-r ]

that's gen formula to work out the sum of...

ok, so..

a= first term = 5
r= common difference = 4.8 / 5 = 0.96

sn is > 124

.:. [ 5(1-0.96^n) ] / [ 1-0.96 ] > 124

.:. [ 5 - (5 x 0.96^n ] / 0.04 > 124

.:. [ 5 - (5 x 0.96^n ] > 4.96

.:. 0.04 > (5 x 0.96^n ]

.:. 0.008 > 0.96^n

ok, so part twooo = find n

so,

log10 (0.008) > log10 (0.96^n)
log10 (0.008) > n x log10 (0.96)
[ log10 (0.008) ] / [ log10 (0.96) ] > n

119 > n

Reply 2

RTJohnson92

just the last part I dunno how to prove it :s-smilie:

9) (iii) The point P on the curve y=(1/2)^x has y-coordinate equal to 1/6. Prove that the x-coordinate of P may be written as:

1 + (log103/log102)

Reply 3

RTJohnson92

This is a 9 mark qu so it must be more complicated then I thought it initially was haha

9) The graph of y= 1 - 3x^-1/2 intersects the x-axis at (9,0)

The shaded region is enclosed by the curve, the x axis and the line x=a(where a>9). Given that the area of the shaded region is 4 square units, find a!!?

HELPPPP :eek:

integrate!!!!

I was doing this paper before. Integrate to get 1/root9 i think. X value is 9, plug that into the integral and it's all good :smile:

Reply 6

RTJohnson92

thanks so much mate, at last someone's explained it step by step :biggrin:

Reply 7

RTJohnson92

I'll have a go but I'll probs still be confused LOLL

Reply 8

hey_its_nay

1/6=(1/2)^x take logs to the base 10 of both sides (not gonna show the 10 for simplicity)
log(1/6)=log(1/2^x)
log1 - log6=xlog1/2
log1 - log6=xlog1 - xlog2 ........ log1=0
-log6=-xlog2 ............... multiply by -1
log6=xlog2
x=log6 / log2 .......... log6=log3+log2
x=log3+log2 / log2
x=log3/log2 + log2/log2
x=log3/log2 +1

all logs are to the base 10