Question 927582
I'll do part a) to get you started


The first term in "26 + 20 + 14 +....+ tn" is *[Tex \LARGE t_{1} = 26] and the common difference is *[Tex \LARGE d = -6] since we're decreasing by 6 each time.



*[Tex \LARGE t_{n} = t_{1}+d(n-1)]



*[Tex \LARGE t_{n} = 26+(-6)(n-1)]



*[Tex \LARGE t_{n} = 26+(-6)(n)+(-6)(-1)]



*[Tex \LARGE t_{n} = 26-6n+6]



*[Tex \LARGE t_{n} = -6n+32]



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*[Tex \LARGE 26 + 20 + 14 +\ldots + t_{n} = \sum_{i=1}^{n}\left(t_{n}\right)]



*[Tex \LARGE 26 + 20 + 14 +\ldots + t_{n} = \frac{n*(t_{1}+t_{n})}{2}]



*[Tex \LARGE 26 + 20 + 14 +\ldots + t_{n} = \frac{n*(26+t_{n})}{2}]



*[Tex \LARGE 26 + 20 + 14 +\ldots + t_{n} = \frac{n*(26+(-6n+32))}{2}]



*[Tex \LARGE 26 + 20 + 14 +\ldots + t_{n} = \frac{n*(26-6n+32)}{2}]



*[Tex \LARGE 26 + 20 + 14 +\ldots + t_{n} = \frac{n*(-6n+58)}{2}]



*[Tex \LARGE -240 = \frac{n*(-6n+58)}{2}]



*[Tex \LARGE -240*2 = n*(-6n+58)]



*[Tex \LARGE -480 = n*(-6n+58)]



*[Tex \LARGE -480 = -6n^2+58n]



*[Tex \LARGE 0 = -6n^2+58n+480]



*[Tex \LARGE -6n^2+58n+480 = 0]



Use the quadratic formula (I'm skipping showing these steps. Let me know if you need me to show these steps) to get



*[Tex \LARGE n = -\frac{16}{3} \ \text{or} \ n = 15]



Ignore any negative or fractional solutions. The only solution is *[Tex \LARGE n = 15]



So there are <font size=6 color="red">15</font> terms if *[Tex \LARGE 26 + 20 + 14 +\ldots + t_{n} = -240]

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