Question

Find the 10th term of the AP where sum of the first n terms is given by 2n² + 3n.​

Answer 1

Answer:

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Step-by-step explanation:

$Given \: , Sn = 2n²+3n$

$now \: we \: know \: that,$

$S1 = a1$

$[ By \: definition \: that \: sum \\ \: of \: first \: term \: is \: a1 \: only]$

$S1 = 2(1)² + 3(1)$

$= 2+3$

$= 5 = a1$

$now \: S2 = 2(2)²+3(2)$

$= 8+6$

$= 14$

$a2 = S2-S1$

$= 14-5$

$= 9 = a2$

$d = a2 - a1$

$= 9-5$

$= 4$

$now \: we \: know \: that ,$

$a10 = a1+ (n-1)d$

$= 5 + (10-1)4$

$= 5+36$

$= 41$

$therefore , \: the \: 10th \\ \: term \: of \: the \: AP \: is \: 41.$

$please \: like \: my \: answer \\ \&\\ mark \: me \: brainliest \: please \: :)))$

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