Help help help!!?

Question

I don't know what wrong i've done,I can't get the right answer for these.

1) Use the expansion of (2-x)^5 to evaluate (1.98)^5 .

2) The first term of an arithmetic progression is a and the common difference is d. If the 5th, 9th and 16th term form a three term geometric progression with common ratio r, find the value of d in terms of a.

Oops sorry.Please

32 - 80x + 80x^2 - 40x^3 + 10x^4 - x^5

(1.98)^5 = (1+0.98)^5
(2-x)^5 = (1+0.98)^5
= 32 - 80(0.98) + 80(0.98)^2
- 40(0.98)^3 + 10(0.98)^4
- (0.98)^5
= 0.08
=

Answer 1

*
You've expanded (2-x)^5 correctly.
Now 1.98 = 2 - .02, so simply plug in .02 into
your expansion:
32 - 80x + 80x^2 - 40x^3 + 10x^4 - x^5
We get
32 - 80(.02) + 80(.0004)-40(.02)³ + 10(.02)^4 - (.02)^5
which works out to
30.4316815968.

2). The 5th term is a + 4d, the 9th term is a + 8d
and the 16th term is a + 15d.
Then we have
a + 8d = r(a+4d)
a + 15d = r²(a+4d).
So
r = (a+8d)/(a+4d) = (a+15d)/(a+8d).
(a+8d)²= (a+4d)(a+15d)
a²+16d+64d² = a²+19d+60d².
This gives
4d²- 3ad=0.
Since d isn't 0,
d = 3a/4.
For an example of such a progression
take a = 4, d = 3.
The 5 term is 16
the 9th term is 28
and
the 16th term is 49.
These 3 terms are in geometric progression
with r = 7/4.

Answer 2

Post your solution and we will tell you where you went wrong.

2 - x = 1. 98 => x = 0.02, so replace x = 0.02 in the expansion.
Your expansion is:
32 - 5(16)(x) + 10(8)(x^2) - 10(2^2)(x^3) + 5( 2) x^4 - x^5.

Usually in those questions, you will be asked for a certain degree of accuracy, so you do not need all the terms.
32 - 80x + 80x^2 - 40x^3 + 10x^4 - x^5.

Substitute 0.02
32 - 80(0.02) + 80(0.02)^2 - 40(0.02)^3 + ...
YOU WILL GET 30.4316816

2) The 5th, 9th and 16th terms are respectively:
a + 4d, a + 8d, a + 15d
Since they form a GP
(a + 8d) / (a + 4d) = (a + 15d)/ (a + 8d)
find a in terms of d from this expression.

Answer 3

geeze!!! you can at least say PLEASE!!!