S3=31 and S6 =3906 then find a and r
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Answered by
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a + ar + ar^2= 31= a(1+r+r^2)
a + ar^+ ar^2+ar^3+ar^4 +ar^5 = 3906= a{1+r +r^2+r^3+r^4+r^5}
s6 = a+ar+ar^2 + a (r^3+r^4+r^5)=3906
31+a(r^3+r^4+r^5)=3906
a(r^3+r^4+r^5)= 3675
r^3a(1+r+r^2)= 3675
r^3 31 =3675
r^3=3675/31
r^3=125
r=5
now s3=31
a(1+r+r^2=31
a(1+5+25)=31
a=1 hope u got it
a + ar^+ ar^2+ar^3+ar^4 +ar^5 = 3906= a{1+r +r^2+r^3+r^4+r^5}
s6 = a+ar+ar^2 + a (r^3+r^4+r^5)=3906
31+a(r^3+r^4+r^5)=3906
a(r^3+r^4+r^5)= 3675
r^3a(1+r+r^2)= 3675
r^3 31 =3675
r^3=3675/31
r^3=125
r=5
now s3=31
a(1+r+r^2=31
a(1+5+25)=31
a=1 hope u got it
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0
Answer:
no
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