Question

Evaluate the given arithmetic series.
7) 9 + 15 + 21 + 27..., n = 14

Answer 1

You just add them

Answer 2

but it does not get 48 when add

Answer 3

Don't include the 14

Answer 4

We're given an arithmetic series with a first term of 9 and a common difference of 6. We need to find the value of the 14th term. We can use the following formula to find the nth term of the series: a_n = a_1 + (n - 1) * d where a_n is the nth term of the sequence, a_1 is the first term of the sequence and d is the common difference. So, substituting the given values, we can get: a_14 = 9 + (14 - 1) * 6 a_14 = 9 + 13 * 6 a_14 = 9 + 78 a_14 = 87 Therefore, the 14th term of the arithmetic series is 87.


If you don't include the 14:


We'd need to determine the value of n, which represents the number of terms in the series. To find the value of n, we can use the formula for the nth term of an arithmetic sequence: an = a1 + (n-1)d Here, a1 is the first term of the sequence (which is 9), d is the common difference between consecutive terms (which is 6), and an is the nth term of the sequence (which we do not know). We can solve for n by plugging these values into the formula and simplifying: an = 9 + (n-1)6 an = 9 + 6n - 6 an = 3 + 6n Now, we can use the given series to find the value of n. We see that the series starts with 9 and increases by 6 each time, so we can write: 9 + 6 + 6(2) + 6(3) + ... + 6(n-1) = an Simplifying this expression gives: 3 + 6 + 9 + 12 + ... + 3n = an This is a sum of an arithmetic series with first term 3, common difference 3, and n terms. Using the formula for the sum of an arithmetic series, we can write: S_n = n/2 * (2a1 + (n-1)d) Substituting our values of a1 = 3, d = 3, and S_n = 9 + 15 + 21 + 27 + ... + (3n) gives: n/2 * (2(3) + 3(n-1)) = 3n Simplifying and solving for n gives: n/2 * (3n + 3) = 3n n^2 + n = 6n n^2 - 5n = 0 n(n - 5) = 0 Therefore, n = 0 or n = 5. However, we know that n is the number of terms in the series and cannot be 0, so n = 5. Therefore, the given arithmetic series has 5 terms.