Question 978886
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You will need these formulas:

{{{a[n]=a[1]+(n-1)d}}} <--nth term of an A.P.
{{{a[n]=a[1]r^(n-1)}}} <--nth term of an G.P.
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The first,fifth and seventh term of an Arthimetic Progression(A.P)...
The first term...is 56, the common difference of the A.P IS d  
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first term of A.P. = {{{a[1]}}} = 56
fifth term of A.P. = {{{a[5]=a[1]+(5-1)d=56+4d}}}
seventh term of A.P. = {{{a[7]=a[1]+(7-1)d=56+6d}}}
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...the first three consecutive terms of a decreasing Geometric Progression(G.P). 
The first term...is 56,...
and the common ratio of the G.P  is r
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first term of G.P. = {{{a[1]}}} = 56
second term of G.P. = {{{a[2]=a[1]r^(2-1)d=56r^1=56r}}}
third term of G.P. = {{{a[3]=a[1]r^(3-1)d=56r^2}}}
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a)(i)Write two equations involving d and r

The first,fifth and seventh term of an Arithmetic Progression(A.P) are
respectively equal to the first three consecutive terms of a decreasing
Geometric Progression(G.P). 
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{{{system(56+4d=56r,56+6d=56r^2)}}}
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(ii)Find the values of d and r
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Divide the first equation through by 4 and the second through by 2
{{{system(14+d=14r,28+3d=28r^2)}}}

Solve the first equation for d  {{{d=14r-14}}}
Substitute it in the second equation:

{{{28+3d=28r^2)}}}
{{{28+3(14r-14)=28r^2)}}}

Solve that quadratic for r.  [You will get two solutions for r, ignore r=1,
because to have a decreasing G.P., r must be a fraction less than 1.

Substitute to find d

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(b) find the sum of the first 10 terms of:
   (I) The arithmetic progression
   (ii) The geometric progression
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You will need these formulas:

{{{S[n]=expr(n/2)(2a[1]+(n-1)^""d)}}} <--sum of first n terms of an A.P.
{{{S[n]=a[1](1-r^n)/(1-r)}}} <--sum of first n terms of a G.P.

Now you can do the problem.  

If you get stuck, tell me in the thank-you
note form below, and I'll get back to you by email.

Edwin</pre>