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Sagot :
Given:
The sum of two terms of GP is 6 and that of first four terms is [tex]\dfrac{15}{2}.[/tex]
To find:
The sum of first six terms.
Solution:
We have,
[tex]S_2=6[/tex]
[tex]S_4=\dfrac{15}{2}[/tex]
Sum of first n terms of a GP is
[tex]S_n=\dfrac{a(1-r^n)}{1-r}[/tex] ...(i)
Putting n=2, we get
[tex]S_2=\dfrac{a(1-r^2)}{1-r}[/tex]
[tex]6=\dfrac{a(1-r)(1+r)}{1-r}[/tex]
[tex]6=a(1+r)[/tex] ...(ii)
Putting n=4, we get
[tex]S_4=\dfrac{a(1-r^4)}{1-r}[/tex]
[tex]\dfrac{15}{2}=\dfrac{a(1-r^2)(1+r^2)}{1-r}[/tex]
[tex]\dfrac{15}{2}=\dfrac{a(1+r)(1-r)(1+r^2)}{1-r}[/tex]
[tex]\dfrac{15}{2}=6(1+r^2)[/tex] (Using (ii))
Divide both sides by 6.
[tex]\dfrac{15}{12}=(1+r^2)[/tex]
[tex]\dfrac{5}{4}-1=r^2[/tex]
[tex]\dfrac{5-4}{4}=r^2[/tex]
[tex]\dfrac{1}{4}=r^2[/tex]
Taking square root on both sides, we get
[tex]\pm \sqrt{\dfrac{1}{4}}=r[/tex]
[tex]\pm \dfrac{1}{2}=r[/tex]
[tex]\pm 0.5=r[/tex]
Case 1: If r is positive, then using (ii) we get
[tex]6=a(1+0.5)[/tex]
[tex]6=a(1.5)[/tex]
[tex]\dfrac{6}{1.5}=a[/tex]
[tex]4=a[/tex]
The sum of first 6 terms is
[tex]S_6=\dfrac{4(1-(0.5)^6)}{(1-0.5)}[/tex]
[tex]S_6=\dfrac{4(1-0.015625)}{0.5}[/tex]
[tex]S_6=8(0.984375)[/tex]
[tex]S_6=7.875[/tex]
Case 2: If r is negative, then using (ii) we get
[tex]6=a(1-0.5)[/tex]
[tex]6=a(0.5)[/tex]
[tex]\dfrac{6}{0.5}=a[/tex]
[tex]12=a[/tex]
The sum of first 6 terms is
[tex]S_6=\dfrac{12(1-(-0.5)^6)}{(1+0.5)}[/tex]
[tex]S_6=\dfrac{12(1-0.015625)}{1.5}[/tex]
[tex]S_6=8(0.984375)[/tex]
[tex]S_6=7.875[/tex]
Therefore, the sum of the first six terms is 7.875.
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