What is the sum of a 30-term arithmetic sequence where the first term is 74 and the last term is -100?
@aum
-468 -442 -416 -390
The common difference d is found as follows: Let the first term be a. 74 = a .....................(1) -100 = a + 29d........(2) If you subtract equation (2) form equation (1), the a terms will cancel and you will have a third equation that you can solve to find the value of d. The sum S of n terms of an arithmetic sequence is found from the formula: \[S=\frac{n}{2}[2a+(n-1)d]\] When you have found the value of d, just plug the d value and the value of n (n = 30 ) and a (a = 74 ) into the formula to find the required sum of the sequence.
subtract equation (2) from equation (1)*
howd you get 29 @kropot72
The first term in the sequence (74) stands alone. The other 29 terms in the sequence are each the result of adding multiples of the common difference to the first term. So the last term in the sequence, the thirtieth, is the first term plus 29 times the common difference.
ohh
Have you calculated the value of common difference d yet?
idk if im right but i got 3
What is the result when you subtract equation (2) from equation (1)? 74 = a .....................(1) -100 = a + 29d........(2)
im so confused
74 - (-100) = a - a -29d ..........(3) Now you need to simplify equation (3)
Taking the left hand side of equation (3), what is: 74 - (-100) = ?
174
Good work! Now looking at the right hand side of equation (3), what is: a - a - 29d = ?
-29d
Correct again! So now we have this equation derived from equation (3): 174 = -29d ...........(4) Now you can solve for d by dividing both sides of equation (4) by -29.
-6
Good work. So we now have d = -6, n = 30, and a = 74 to plug into the formula for the sum, giving: \[S=\frac{30}{2}[(74\times2)+(29\times -6)]=you\ can\ calculate\]
-390
Good work! You are correct.
thanks
You're welcome :)
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