explain it too fast.
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Given that,
t1=a
t2=b
t3=c
common difference d =t2-t1
=b-a
tn =a+(n-1)d
c=a+(n-1)(b-a)
n=b+c-2a/b-a
Hence,
Sn=n/2(a+l)
=(a+c) (b+c-2a) /2(b-a) Hence proven.
Hope this helps you.....
t1=a
t2=b
t3=c
common difference d =t2-t1
=b-a
tn =a+(n-1)d
c=a+(n-1)(b-a)
n=b+c-2a/b-a
Hence,
Sn=n/2(a+l)
=(a+c) (b+c-2a) /2(b-a) Hence proven.
Hope this helps you.....
kriyamaths:
please mark my answer as brainliest...
thank you!
okk.
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