Find the area of the triangle.
1 answer:
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Hmm if I'm not mistaken, is just an "ordinary" annuity, thus
![\bf \qquad \qquad \textit{Future Value of an ordinary annuity} \\\\ A=pymnt\left[ \cfrac{\left( 1+\frac{r}{n} \right)^{nt}-1}{\frac{r}{n}} \right] \\\\\\](https://tex.z-dn.net/?f=%5Cbf%20%5Cqquad%20%5Cqquad%20%5Ctextit%7BFuture%20Value%20of%20an%20ordinary%20annuity%7D%0A%5C%5C%5C%5C%0AA%3Dpymnt%5Cleft%5B%20%5Ccfrac%7B%5Cleft%28%201%2B%5Cfrac%7Br%7D%7Bn%7D%20%5Cright%29%5E%7Bnt%7D-1%7D%7B%5Cfrac%7Br%7D%7Bn%7D%7D%20%5Cright%5D%0A%5C%5C%5C%5C%5C%5C)
![\bf \begin{cases} A= \begin{array}{llll} \textit{original amount}\\ \textit{already compounded} \end{array} & \begin{array}{llll} \end{array}\\ pymnt=\textit{periodic payments}\to &400\\ r=rate\to 10\%\to \frac{10}{100}\to &0.1\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, so once} \end{array}\to &1\\ t=years\to &14 \end{cases} \\\\\\ A=400\left[ \cfrac{\left( 1+\frac{0.1}{1} \right)^{1\cdot 14}-1}{\frac{0.1}{1}} \right] ](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%0AA%3D%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A%5Ctextit%7Boriginal%20amount%7D%5C%5C%0A%5Ctextit%7Balready%20compounded%7D%0A%5Cend%7Barray%7D%20%26%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A%0A%5Cend%7Barray%7D%5C%5C%0Apymnt%3D%5Ctextit%7Bperiodic%20payments%7D%5Cto%20%26400%5C%5C%0Ar%3Drate%5Cto%2010%5C%25%5Cto%20%5Cfrac%7B10%7D%7B100%7D%5Cto%20%260.1%5C%5C%0An%3D%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A%5Ctextit%7Btimes%20it%20compounds%20per%20year%7D%5C%5C%0A%5Ctextit%7Bannually%2C%20so%20once%7D%0A%5Cend%7Barray%7D%5Cto%20%261%5C%5C%0A%0At%3Dyears%5Cto%20%2614%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0AA%3D400%5Cleft%5B%20%5Ccfrac%7B%5Cleft%28%201%2B%5Cfrac%7B0.1%7D%7B1%7D%20%5Cright%29%5E%7B1%5Ccdot%2014%7D-1%7D%7B%5Cfrac%7B0.1%7D%7B1%7D%7D%20%5Cright%5D%0A)
We have to determine the value of
= (5(9)+6) + (5(10)+6) +(5(11)+6) + .......... + (5(21)+6)
= 51+56+61+66+ ........ + 111
Since, the common difference is 5, hence this series is in arithmetic progression.
Sum of AP is given by the formula:
Since, there are 13 terms.
=
=
=
=
= 1053
Therefore, the sum of the series is 1053.
So, Option G is the correct answer.