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Question

Question 29
Find the sum of all the 11 terms of an AP whose middle most term is 30.

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Solution

Total number of terms (n) = 11 [odd]
Middle most term=(n+1)2th term=(11+12)th term=6th term

Hence, a6=30
a+(61)d=30 [ an=a+(n1)d]
a+5d=30...(i)

Sum of terms of an AP,
Sn=n2[2a+(n1)d]
S11=112[2a+(n1)d]
=112(2a+10d)=11(a+5d)
=11×30=330 [from eq(i)]

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