The sum of first n terms of a AP is given by Sn=2n.n+3n find the sixteenth term of AP
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Let an be the nth term & Sn be the sum of first n terms.
Given:
Sn = 2n²+3n
an = 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 +1
an = 4n+1
a16= 4×16 +1= 64+1= 65
Hence, the 16th term of an AP is 65
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Hope this will help you...
Given:
Sn = 2n²+3n
an = 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 +1
an = 4n+1
a16= 4×16 +1= 64+1= 65
Hence, the 16th term of an AP is 65
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Hope this will help you...
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