How do you find the first three terms of the arithmetic series n=31, $a_{n} = 78$ $a_n=78$ , $S_{n} = 1023$ $S_n=1023$ ?

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How do you find the first three terms of the arithmetic series n=31, $a_{n} = 78$ $a_n=78$ , $S_{n} = 1023$ $S_n=1023$ ?

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Answer 1 · 2017-08-26T09:14:06

First three terms are $- 12$ $-12$ , $- 9$ $-9$ and $- 6$ $-6$

Explanation:

In an arithmetic series with $a_{1}$ $a_1$ as first term and $d$ $d$ as common difference, while $n^{t h}$ $n^(th)$ term $a_{n} = a_{1} + (n - 1) \times d$ $a_n=a_1+(n-1)xxd$ , the sum of series up to $n^{t h}$ $n^(th)$ terms is $S_{n} = \frac{n}{2} (a_{1} + a_{n}) = \frac{n}{2} (2 a + (n - 1) d)$ $S_n=n/2(a_1+a_n)=n/2(2a+(n-1)d)$ .

Here we have $n = 31$ $n=31$ , $a_{n} = a_{1} + 30 d = 78$ $a_n=a_1+30d=78$

and $S_{n} = 1023 = \frac{31}{2} (a_{1} + 78)$ $S_n=1023=31/2(a_1+78)$

Hence $a_{1} = 1023 \times \frac{2}{31} - 78 = 33 \times 2 - 78 = - 12$ $a_1=1023xx2/31-78=33xx2-78=-12$

and as $78 = - 12 + 30 \times d$ $78=-12+30xxd$ we have $d = \frac{78 + 12}{30} = \frac{90}{30} = 3$ $d=(78+12)/30=90/30=3$

Hence, first three terms are $- 12$ $-12$ , $- 12 + 3 = - 9$ $-12+3=-9$ and $- 9 + 3 = - 6$ $-9+3=-6$

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How do you find the first three terms of the arithmetic series n=31, $a_{n} = 78$ $a_n=78$ , $S_{n} = 1023$ $S_n=1023$ ?

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Explanation:

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How do you find the first three terms of the arithmetic series n=31, an=78a_n=78, Sn=1023S_n=1023?

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How do you find the first three terms of the arithmetic series n=31, $a_{n} = 78$ $a_n=78$ , $S_{n} = 1023$ $S_n=1023$ ?