What is the sum of the arithmetic series where n=50 , t1=9 and t2=331
To find the sum of an arithmetic series, we need to use the formula:
Sn = (n/2)(t1 + tn)
where Sn is the sum of the series, n is the number of terms, t1 is the first term, and tn is the nth term.
In this case, n = 50, t1 = 9, and t2 = 331. We need to find tn, the 50th term.
To find the common difference (d), we can subtract t1 from t2:
d = t2 - t1
= 331 - 9
= 322
Now we can find tn using the formula:
tn = t1 + (n-1)d
tn = 9 + (50-1) * 322
= 9 + 49 * 322
= 9 + 15678
= 15687
Finally, we can substitute all the values into the sum formula:
Sn = (n/2)(t1 + tn)
= (50/2)(9 + 15687)
= (25)(15696)
= 392,400
Therefore, the sum of the arithmetic series where n = 50, t1 = 9, and t2 = 331 is 392,400.
To find the sum of an arithmetic series, you can use the formula:
Sn = (n/2) * (t1 + tn)
where Sn is the sum of the series, n is the number of terms, t1 is the first term, and tn is the nth term.
In this case, n is given as 50, t1 is given as 9, and t2 is given as 331. We need to find tn.
To find tn, we can use the formula for the nth term of an arithmetic series:
tn = t1 + (n-1)d
where d is the common difference.
To find d, we can subtract t1 from t2 and divide it by 1, since t2 is the second term in the series:
d = (t2 - t1) / 1
= (331 - 9) / 1
= 322
Now, we can substitute the values of n, t1, tn, and d into the sum formula:
Sn = (n/2) * (t1 + tn)
= (50/2) * (9 + (t1 + (n-1)d))
= 25 * (9 + (9 + (50-1) * 322))
= 25 * (9 + (9 + 49 * 322))
= 25 * (9 + 9 + 15778)
= 25 * 15796
= 394900
Therefore, the sum of the arithmetic series is 394900.
This is exactly the same type of question as your previous one.
Just follow my steps using these numbers.