The sum of first n terms of an AP is given by Find the sixteenth termo

1.	The sum of first n terms of an AP is given by Find the sixteenth termof the AP.
Answer» Given ,\xa0{tex}{S_n} = 5{n^2} - 3n{/tex}Now,\xa0{tex}{a_n} = {S_n} - {S_{n - 1}}{/tex}{tex} = 5{n^2} - 3n - [5{(n - 1)^2} - 3(n - 1)]{/tex}{tex} \\Rightarrow {a_n} = 10n - 8{/tex} ............ (i)Clearly,\xa0{tex}{a_{16}} = 10 \\times 16 - 8 = 152{/tex}Now, for finding AP, put n = 1, 2, 3, 4 ... in Eq(i).\xa0So, from Eq(i), we have{tex}{a_{1}} = 2{/tex}{tex}{a_{2}} =1 2{/tex}{tex}{a_{3}} = 22{/tex}Hence, The AP is 2,12, 22....Aliter:{tex}a_1=s_1=5(1)^2-3(1)=2{/tex}{tex}a_2=s_2-s_1=[5(2)^2-3(2)]-[5(1)^2-3(1)]=12{/tex}{tex}d=a_2-a_1=12-2=10{/tex}{tex}\\therefore{/tex}\xa0AP is 2,12,22...16th term is :{tex}a_{16}=2+15\\times10=152{/tex}

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