Question

find number of terms....

Answer 1

First term (a) = 5
Last term (tn ) = 45
Sum (Sn) = 400
$sum = \frac{n}{2} (a + \: tn)$
$400 = \frac{n}{2} (5 + 45)$
800 = n (50)
$n = \frac{800}{50}$
n = 16

Tn = a + (n-1)d
45 = 5 + (16-1)d
40=15d
$d = \frac{40}{15}$
$d = \frac{8}{3}$
d = 2.6.......



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Pranshu Gupta

Answer 2

hello users ...
we have given that:-
a = 5
An = 45 
and
Sn = 400 
where,a is the first term , An is the last term of an AP and
Sn is the sum of n terms 

we have to find :
n = ? (number of terms)
and
d=? (common difference )

solution:-
we know that
$S_{n} = \frac{n}{2} [a + A_{n} ]$

and

$A_{n} = a + (n-1)d$

Here,
Sn = 400
=> 400 =  n/2[ 5 + 45]
=> 800 = n*50
=> n = 800/50 = 16 

now,
Last term.
$A_{16} = 5 + (16 -1) d$

=> 45 = 5 + 15d 

=> 40 = 15 d

=> d = 40 / 15 ≈ 2.67 Answer 

@ hope it helps ..