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Question
Find the sum of n terms of an A.P whose nth terms is given by an=5−6n.
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Solution
Given:tn=5−6n
Putn=1,t1=−1
n=2,t2=−7
n=3t3=−13
Sum of n terms=n/2[2a+(n−1)d]=n/2[2(−1)+(n−1)−6]
=n/2[−2−6n+6]
=n/2(4−6n)
=>n(2−3n)
Putn=1,t1=−1
n=2,t2=−7
n=3
t3=−13
Sum of n terms=n/2[2a+(n−1)d]
=n/2[2(−1)+(n−1)−6]
=n/2[−2−6n+6]
=n/2(4−6n)
=>n(2−3n)
=n/2[−2−6n+6]
=n/2(4−6n)
=>n(2−3n)
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