P1 Maths October November 2007 question 4 (iii)

[XWTBSILT]

October November 2007 question 4 (iii) ONLY

The 1st term of an arithmetic progression is a and the common difference is d, where d ? 0.

(i) Write down expressions, in terms of a and d, for the 5th term and the 15th term. [1]
The 1st term, the 5th term and the 15th term of the arithmetic progression are the first three terms of
a geometric progression.
(ii) Show that 3a = 8d. [3]
(iii) Find the common ratio of the geometric progression. [2]
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Please include all ur steps and thankx in advance

 

[XWTBSILT]

Got it

ar/a=r

take the ratio of any 2 terms

eq 1 a+4d/ a or also a+14d/a+4d

substitute either a or d with the other.

3a=8d
a=8/3 d
sub this for eq 1

(8/3d + 4d)/ (8/3d) = 2.5
r=2.5

 

[Sid]

Here is it:
(i) A.P 1st term:a C.D: d
a+4d=5th term
a=14d=15th term
G.P
a=1st term
ar=2nd term
ar^2= 3rd term
Now,
(ii) a=a -(1)
a+4d=ar -(2)
a+14d=ar^2 -(3)
we pick the 2nd equation
r=a+4d/a
equate eq.2 n 3 by subttituing the r=a+4d/a
a+14d=a(a+4d/a)^2
it comes out to be
a^2+14ad=a^2+8ad+16d^2
By solving we get
6ad=16d^2
3a=8d (Shown)
(iii) take d=3a/8
substitute in r=a+4d/a or any eq.u want
r=a+4(3a/8)/a
r=2.5a/a
r=2.5
there it goes.....
i hope u understand my long method