the sum of first n terms of an ap is given by SN = 2 n square 3 n. the 16 term of the ap
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HELLO DEAR,
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
Then,
the 16th term of an AP is 65
I HOPE ITS HELP YOU DEAR,
THANKS
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
Then,
the 16th term of an AP is 65
I HOPE ITS HELP YOU DEAR,
THANKS
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