Answer:
Common ratio=1/2
Sum of first 6 terms of an A.P=31.5
Step-by-step explanation:
Let a be the first term and d be the common difference of an A.P
a=4
nth term of an A.P
[tex]a_n=a+(n-1)d[/tex]
Using the formula
[tex]a_4=a+3d=4+3d[/tex]
[tex]a_{10}=a+9d=4+9d[/tex]
According to question
We know that
For G.P
[tex]\frac{a_n}{a_{n-1}}=r[/tex]=Common ratio
[tex]r=\frac{4+3d}{4+9d}=\frac{4}{4+3d}[/tex]
[tex]\frac{4+3d}{4+9d}=\frac{4}{4+3d}[/tex]
[tex](4+3d)^2=\frac{4}{4+3d}[/tex]
[tex]16+9d^2+24d=16+36d[/tex]
[tex]9d^2+24d-36d-16+16=0[/tex]
[tex]9d^2-12d=0[/tex]
[tex]3d(3d-4)=0[/tex]
[tex]d=0, d=4/3[/tex]
Substitute the value of d
When d=0
[tex]r=\frac{4}{4+3d}=\frac{4}{4+0}=1[/tex]
When d=4/3
[tex]r=\frac{4}{4+3\times \frac{4}{3}}=\frac{1}{2}[/tex]
But we reject d=0 because if we take d=0 then the terms are not consecutive terms of G.P
Sum of n terms of an A.P
[tex]S_n=\frac{n}{2}(2a+(n-1)d)[/tex]
Using the formula
Substitute n=6 and d=1/2
[tex]S_6=\frac{6}{2}(2(4)+5\times \frac{1}{2})[/tex]
[tex]S_6=31.5[/tex]
