Question

The sum of first n terms of an ap is n÷2(3n+13)find its 30 th term

Answer 1

$Hey \: \: there$
$Let \: a \: be \: the \: first \: term \: d \: be \: the \: common difference \: n \: \: be \: \: the number \: of \: terms \\ we \: are \: given \: that \: - \\ \\ Sn \: = \frac{n}{2} \times (3n \: + 13) \\ \\ if \: \: n \: = 1 \: \\ \: then \: \\ \\ \: Sn \: = \frac{1}{2} \times (3 + 13) \\ \\ = > \frac{1}{2} \times 16 \\ \\ = > 8 \\ \\ which \: is \: \: a \: first \: term \: also \: \\ \\ now \: if \: \: Sn \ \: = 2 \\ then \: \: = > \frac{2}{2} \times (3 \times 2 + 13) \\ \\ = 1 \times 19 \\ \\ = 19 \: \: now \: to \: get \: \:$
a₂ = S₂ - S₁

= 19 - 8

= 11

now common difference , d = a₂ - a₁

= 11 - 8

= 3

Now we need to find a₃₀

a₃₀ = a + 29 d

= 8 + 29 × 3

= 8 + 87

= 95.

Hence a₃₀ =95


⨷ Hope this would help you ⨷

Answer 2

Heya User ] =_= [ 

--> $S_n = \frac{n}{2} ( 3n + 13 )$

--> Now, a clever observation would tell uh this --> 
--> [tex]a_n = [ a_1 + a_2 +... + a_{n-1} + a_n ] - [ a_1 + a_2 +... + a_{n-1} ]\ \\ =\ \textgreater \ a_n = S_n - S_{n-1}[/tex]

=> $a_{30} = S_{30}- S_{29} = \frac{30}{2} [ 3 * 30 + 13 ] - \frac{29}{2} [ 3 * 29 + 13 ]\\ =\ \textgreater \ a_{30} = 15 * 103 - 50*29 = 95$

Hence, we get the 30th term as --> 95