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Question

Find the sum of the following arithmetic progressions:
41,36,31,..... to 12 terms.

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Solution

Given A.P is
41,36,31,.....
number of terms of A.P is n=12
first term of this A.P is a1=41
second term of this A.P is a2=36
common difference
d=a2a1=3641=5
we know that sum of n term of an A.P is given by
Sn=n2[2a+(n1)d)]
S12=122[2×41+(121)×5]
S12=6[82+(11)×5)]
S12=6[8255]
S12=6[27]
S12=162
hence the sum of 12 terms of given A.P is 162.

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