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An Exponential Sequence GP has a positive Common ratio. If the Sum of Infinity Sequence is 25 and the Sum of the first two terms is 16, find the (a) 5th term (b) The Sum of the first four terms

Mathematics

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Answers

2 parts
a

162125\frac{162}{125}

b

54425\frac{544}{25}

Step-by-step solution

3 steps

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Part a· 2 steps

The 5th term

  1. Recall the first term and common ratio.

    From the previous calculations, the first term is a=10a=10 and the common ratio is r=35r=\frac{3}{5}.

  2. Calculate the 5th term using the formula Tn=arn−1T_n = ar^{n-1}.

    T5=10(35)5−1T_5 = 10 \left(\frac{3}{5}\right)^{5-1} T5=10(35)4T_5 = 10 \left(\frac{3}{5}\right)^4 T5=10×81625T_5 = 10 \times \frac{81}{625} T5=810625T_5 = \frac{810}{625} T5=162125T_5 = \frac{162}{125}

    The 5th term is 162125\boxed{\frac{162}{125}}.

Part b

54425\frac{544}{25}

The sum of the first four terms

  1. Calculate the sum of the first four terms using the formula Sn=a(1−rn)1−rS_n = \frac{a(1-r^n)}{1-r}.

    S4=10(1−(35)4)1−35S_4 = \frac{10\left(1-\left(\frac{3}{5}\right)^4\right)}{1-\frac{3}{5}} S4=10(1−81625)25S_4 = \frac{10\left(1-\frac{81}{625}\right)}{\frac{2}{5}} S4=10(625−81625)25S_4 = \frac{10\left(\frac{625-81}{625}\right)}{\frac{2}{5}} S4=10(544625)25S_4 = \frac{10\left(\frac{544}{625}\right)}{\frac{2}{5}} S4=10×544625×52S_4 = 10 \times \frac{544}{625} \times \frac{5}{2} S4=5440×51250S_4 = \frac{5440 \times 5}{1250} S4=272001250S_4 = \frac{27200}{1250} S4=54425S_4 = \frac{544}{25}

    The sum of the first four terms is 54425\boxed{\frac{544}{25}}.

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