The first term of an arithmetic sequence is -12 and the last term is 40. if the sum of the series is 196, find the number of terms in the sequence and the common difference.


The only formulas I've gotten are Sn=n/2[2a+(n-1)d] and Tn=a+(n-1) where in both formulas a=the first term and d=the c common difference.

Any help will be much appreciated. Thanks again.​

Answers

Answer 1

The number of terms in the arithmetic sequence is 14, and the common difference is 8.

Let's use the given formulas to solve the problem. We are given that the first term (a) is -12 and the last term (Tn) is 40. We also know that the sum of the series (Sn) is 196.

Using the formula for the sum of an arithmetic series (Sn = n/2[2a + (n-1)d]), we can substitute the given values into the equation and solve for n:

196 = n/2[-24 + (n-1)d]

Next, let's use the formula for the nth term of an arithmetic sequence (Tn = a + (n-1)d):

40 = -12 + (n-1)d

Now we have a system of two equations:

196 = n/2[-24 + (n-1)d]

40 = -12 + (n-1)d

We can solve this system of equations to find the values of n and d. By substituting the value of d from the second equation into the first equation, we get:

196 = n/2[-24 + 52 - 4d]

Simplifying further:

196 = n/2[28 - 4d]

Multiplying both sides by 2 to eliminate the fraction:

392 = n[28 - 4d]

Now, let's substitute the value of n in the second equation:

40 = -12 + 14d

Rearranging the equation:

14d = 52

Solving for d:

d = 52/14

d ≈ 3.714

Substituting this value of d back into the second equation, we can find n:

40 = -12 + 14(3.714)

40 = -12 + 51.996

40 ≈ 39.996

Since the number of terms (n) must be a whole number, we can round 39.996 to the nearest whole number, which is 40.

Therefore, the number of terms in the sequence is 40, and the common difference is approximately 3.714.

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Complete question: The first and last term of an Arithmetic progression is -12 and 40 if the sum of the sequence is 196 find the number of terms and the common difference.


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