Sunday

December 21, 2014

December 21, 2014

Posted by **Anonymous** on Sunday, February 21, 2010 at 8:10am.

A)If 'a' is the first number and 'd' the common difference, show that a+d=8

B)If the first number is decreased by 1 and the second number by 2, the three numbers then form a geometric sequence. Find the three terms.

- maths -
**Reiny**, Sunday, February 21, 2010 at 8:39amyour last 3 postings all deal with the basic definitions of sequences and series and involve translating the English statements into mathematical expressions.

Why don't you show me what you have done for this one so far, and I will gladly help you further.

- maths -
**Anonymous**, Sunday, February 21, 2010 at 8:49amok cool... a) S/n=n/2 [2a+(n-1)d]

24=3/2[2a+2d]

48=3[2a+2d]

16=2a+2d

8=a+d

B)i dont know from here hey...

my name is marco

- maths -
**Reiny**, Sunday, February 21, 2010 at 9:02amok Marko, it would have been easier to just do this

a + a+d + a+2d = 24

3a + 3d = 24

a + d = 8

b) so now we know that a = 8-d

"If the first number is decreased by 1" so term1 = a-1

" the second number by 2" , so term2 = a+d - 2

our new terms are then

t_{1}= a-1 = 8-d - 1 = 7-d

t_{2}= a+d-2 = 8-d-d-2 = 6

t_{3}= a+2d = 8-d + 2d = 8+d

now these form a GS

then

6/(7-d) = (8+d)/6

take over.

- correction - maths -
**Reiny**, Sunday, February 21, 2010 at 9:05amnear the end I had a typo but it did not cause an error

t_{2 = a+d-2 = 8-d-d-2 = 6 should have been t2 = a+d-2 = 8-d+d-2 = 6 } - maths -
**Anonymous**, Sunday, February 21, 2010 at 9:20amok thx.. i got 36=(8+d)(7-d)

36= 56-8d+7d-d squared

36=-d^2-d+56

d^2+d-56+36=0

d^2+d-20=0

(d-4)(d+5)=o

d=4 d=-5

therefore

T1=3

T2=6

T3=12

and

T1=12

T2=6

T3=3

- maths -
**Reiny**, Sunday, February 21, 2010 at 9:38amNice, see how easy they are ?

- maths -
**Anonymous**, Sunday, February 21, 2010 at 10:13amya thx a ton

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