1.

Find the sum of first 51 terms of an A.P.whose second and third terms are 14 and 18 respectively

Answer» \xa0given,a2=14\xa0a3=18\xa0d=a3−a2=4\xa0a1=a2−d\xa0a1 =14-4a=10Sn=n/2(2a+(n-1)d)\xa0Sn=51/2(2*10+(51-1)4)\xa0Sn=5610


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