Question

if an ap the first term is 8,Nth term is 33 and sum of 1st n term is 123 find n and d

Answer 1

a=8
Tn=33
Sn=123
$tn = a + (n - 1)d \\ 33 = 8 + (n - 1)d \\ \frac{25}{n - 1} = d \\ sn = \frac{n}{2} (2a + (n - 1)d \\ 123 = \frac{n}{2} (2 \times 8 + (n - 1)( \frac{25}{n - 1} ) \\ 123= \frac{n}{2} (16 + 25) \\ 123 \times \frac{2}{41} = n \\ n = 6 \\ tn = 33 \\ 33 = 8 + (6 - 1)d \\ 25 \times \frac{1}{5} = d \\ d = 5$
nth term is 6
and common difference is 5.
don't forget to do it brainleist.

Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Assumption

Common difference be d

Then,

8 + (n - 1)d = 33

(n - 1)d = 33 - 8

(n - 1)d = 25 ..... (1)

Now,

$\tt{\rightarrow\dfrac{n}{2}[16+(n-1)d]=123}$

Putting value of (1)

$\tt{\rightarrow\dfrac{n}{2}[16+25]=123}$

$\tt{\rightarrow n=\dfrac{123\times 2}{41}}$

$\tt{\rightarrow n=\dfrac{123\times 2}{41}}$

$\tt{\rightarrow n=\dfrac{246}{41}}$

= 6

Hence,

From (1)

(6 - 1)d = 25

5d = 25

$\tt{\rightarrow d=\dfrac{25}{5}}$

d = 5

Hence we get

d = 5 and n = 6