Find the number of Term of the AP -12,-9,-6.......21.If 1 is added to each term of this AP then find the sum of all the terms of the AP thus obtained.
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Answer:
tn=a+(n-1)d
21=-12+(n-1)3
21=-12+3n-3
21=-15+3n
21+15=3n
36=3n
36/3=n
n=12
Sn=n/2 (2a+(n-1)d)
=12/2 (2×(-12)+(21-1)3)
=6(-24+33)
=6×9
=54.
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