find number of terms....
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First term (a) = 5
Last term (tn ) = 45
Sum (Sn) = 400


800 = n (50)

n = 16
Tn = a + (n-1)d
45 = 5 + (16-1)d
40=15d


d = 2.6.......
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Pranshu Gupta
Last term (tn ) = 45
Sum (Sn) = 400
800 = n (50)
n = 16
Tn = a + (n-1)d
45 = 5 + (16-1)d
40=15d
d = 2.6.......
Mark as brainliest answer.
Follow me
Pranshu Gupta
PranshuGupta1:
mark as brainliest
Answered by
1
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} ] S_{n} = \frac{n}{2} [a + A_{n} ]](https://tex.z-dn.net/?f=+S_%7Bn%7D+%3D++%5Cfrac%7Bn%7D%7B2%7D+%5Ba+%2B++A_%7Bn%7D+%5D)
and

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

=> 45 = 5 + 15d
=> 40 = 15 d
=> d = 40 / 15 ≈ 2.67 Answer
@ hope it helps ..
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
and
Here,
Sn = 400
=> 400 = n/2[ 5 + 45]
=> 800 = n*50
=> n = 800/50 = 16
now,
Last term.
=> 45 = 5 + 15d
=> 40 = 15 d
=> d = 40 / 15 ≈ 2.67 Answer
@ hope it helps ..
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