if the sum of first 3 terms of an ap is equal to the sum first 8 terms than show that 6th term is 0
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Step-by-step explanation:
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and
Let, the first term and common difference of the respective AP be
a
d.
Again, let,
Sum of the first
m
-terms of the AP
=Sm
,
Sum of the first
n
-terms of the AP
=Sn
& Sum of the first
(m+n)
-terms of the AP
=Sm+n.
According to the question,
Sm=Sn
⟹m2{2a+(m−1)d}=n2{2a+(n−1)d}
⟹2am+(m2−m)d=2an+(n2−n)d
⟹2a(m−n)+{(m2−n2)−(m−n)}d=0
⟹(m−n){2a+(m+n−1)d}=0
⟹2a+(m+n−1)d=0...(⋆)
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