What is the sum of first 12 terms of an arithmetic progression, if the first term is -19 and the last term is 36?
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Answer:
given: a=-19
tn=36
n= 12
Step-by-step explanation:
tn = a+(n-1)d
36 = -19+( 12-1)d
36 =-19+11d
36+19=11d
55= 11d
d=5....
sn= n/2( 2a+(n-1)d)
s12 = 12/2 (2×-19+11×5)
=6 (-38+55)
=6×17
=102 is ur answer
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