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The sum of first n terms of an AP is given by S212 +3n. Find the sixteenth term of the AP |
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Answer» Given:Sn = 2n²+3nan = S(n) - S(n-1)=2n²+3n -[ 2(n-1)²+3(n-1)]= 2n²+3n -[2(n²+1-2n)+3n -3]= 2n²+3n -[2n²+2-4n + 3n -3]= 2n²+3n -2n²+4n-3n -2+3= 2n²+2n²+3n -3n +4n -2+3= 4n +1an = 4n+1a16= 4×16 +1=64+1= 65Hence the 16th term is 65 |
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