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if Sn the sum of first n terms of an AP is given by Sn =3n^2-4n,then find its nth term.

Answer» Sn = 3n2 + 4n.a1\xa0= S1 = 3(1)2\xa0+ 4(1) = 7a1\xa0+ a2\xa0= S2\xa0= 3(2)2 + 4(2)= 12 + 8= 20a2 = S2- S1\xa0= 20 - 7 = 13or, {tex}a + d = 13{/tex}or, {tex}7 + d = 13{/tex}{tex}\\therefore{/tex}\xa0{tex}d = 13 - 7 = 6{/tex}{tex}\\therefore{/tex}\xa0A.P. becomes 7,13,19,......\xa0Now, {tex}a_n =a + (n - 1 )d{/tex}{tex}= 7 + (n - 1)(6){/tex}{tex}= 7 + 6n - 6{/tex}{tex}= 6n + 1{/tex}or, an = 6n + 1


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