If the 3rd an 11 term of an AP is 8 and respectively, find the sum of first 10 terms
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a+2d=8 and a+10d=20
⇒a=5,d= 3/2
∴S10 = 10/2. [2a+(10−1)d]
=5[2×5+9×3/2]
=117 1/2
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