Determine the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively
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Answer:
5610
Step-by-step explanation:
Sn=n/2[2a+(n-1)d]
=S51=51/2[2*10+(50)4]
=51/2[20+200]
=51/2 * 220
=51*110 (By cancellation)
=5610
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