Question 1006281
We use {{{a[n]=a[1]+(n-1)d}}}, for the nth term of an A.P.
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>>1.given that the tenth, fourth and first terms of an A.P<< 
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{{{a[10]=a[1]+(10-1)d=4+9d}}}

{{{a[4]=a[1]+(4-1)d=4+3d}}}

{{{a[1]=4}}}
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>>are the three consecutive terms of the G.P.<<
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{{{(4+9d)r = 4+3d}}} or {{{4r+9dr=4+3d}}}

{{{(4+3d)r = 4}}}    or {{{4r+3dr=4}}}

We have the system:

{{{system(4r+9dr=4+3d,4r+3dr=4)}}}

Solve that system by solving one for a variable and
substituting in the other.  (Lot of messy work). 
The two solutions are:

(d,r) = (0,1) or (4/3, 1/2)

The first solution, (d,r)=(0,1) gives the trivial sequence 
4,4,4,4,4,4,4... for both the A.P. with d=0 and a G.P. with r=1.

4, 4, 4, 4, 4, 4, 4, 4, 4, 4 = A.P. with common difference 0
4,       4,                4 = G.P. which common ratio 1...

The second solution (d,r) = (4/3, 1/2) gives the A.P

4, 16/3, 20/3, 8, 28/3, 32/3, 12, 40/3, 44/3, 16 = A.P.
4,             8,                           , 16 = G.P.

The sum of the first 6 terms is given by the sum formula:

{{{S[n]=expr(n/2)(2a[1]+(n-1)d^"")}}}

{{{S[6]=expr(6/2)(2*4+(6-1)(4/3)^"")}}}

{{{S[6]=3(8+5(4/3)^"")}}}

{{{S[6]=3(8+20/3)}}}

{{{S[6]=24+20}}}

{{{S[6]=44}}}

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In the trivial case, 4,4,4,4,4,4,..., the sum of the first
6 terms is 24.

Edwin</pre>