Assume the series 10+18+26... continues for 200 terms. What is the sum?

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Assume the series 10+18+26... continues for 200 terms. What is the sum?

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Answer 1 · 2016-01-25T17:54:44

$a_{2} - a_{1} = 18 - 10 = 8$ $a_2-a_1=18-10=8$
$a_{3} - a_{2} = 26 - 18 = 8$ $a_3-a_2=26-18=8$

$\Rightarrow$ $implies$ This is an arithmetic series .

$\Rightarrow$ $implies$ common difference $= d = 8$ $=d=8$
first term $= a_{1} = 10$ $=a_1=10$

The sum of arithmetic series is given by

$S u m = \frac{n}{2} {2 a_{1} + (n - 1) d}$ $Sum=n/2{2a_1+(n-1)d}$

Where $n$ $n$ is the number of terms, $a_{1}$ $a_1$ is the first term and $d$ $d$ is the common difference.

Here $a_{1} = 10$ $a_1=10$ , $d = 8$ $d=8$ and $n = 200$ $n=200$

$\Rightarrow S u m = \frac{200}{2} {2 \cdot 10 + (200 - 1) 8} = 100 (20 + 199 \cdot 8) = 100 (20 + 1592) = 100 \cdot 1612 = 161200$ $implies Sum=200/2{2*10+(200-1)8}=100(20+199*8)=100(20+1592)=100*1612=161200$

Hence the sum is $161200$ $161200$ .

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Assume the series 10+18+26... continues for 200 terms. What is the sum?

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