Question

the third term of an arithmetic sequence is -8 and the sum of the first 10 terms of the sequence is -230. Find:
a) The first term of the sequence
b) The sum of the first 13 terms.

Answer 1

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Answer 2

Given,

Third term of an arithmetic sequence is -8.

Sum of the first 10 terms of the sequence is -230.

Solution,

Consider the first term of A.P. is a, common difference is d, then nth term,

$\[{a_n} = a + \left( {n - 1} \right)d\]$ and sum of n terms is $\[{S_n} = \frac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)\]$.

$\[\begin{array}{l}a + \left( {3 - 1} \right)d = - 8\\ \Rightarrow a + 2d = - 8\end{array}\]$---------(1)

$\[\begin{array}{l}\frac{{10}}{2}\left( {2a + \left( {10 - 1} \right)d} \right) = - 230\\ \Rightarrow 2a + 9d = - 46\end{array}\]$---------(2)

a) The first term of the sequence.

Put the value a from equation 1 into 2,

$\[\begin{array}{l} \Rightarrow 2\left( { - 8 - 2d} \right) + 9d = - 46\\ \Rightarrow - 16 - 4d + 9d = - 46\\ \Rightarrow 5d = - 30\\ \Rightarrow d = - 6\end{array}\]$

$\[\begin{array}{l}a + 2d = - 8\\ \Rightarrow a + 2 \times \left( { - 6} \right) = - 8\\ \Rightarrow a - 12 = - 8\\ \Rightarrow a = 4\end{array}\]$

Hence the first term is 4.

b) The sum of the first 13 terms.

$\[\begin{array}{l}{S_{13}} = \frac{{13}}{2}\left( {2a + \left( {13 - 1} \right)d} \right)\\ \Rightarrow {S_{13}} = \frac{{13}}{2}\left( {2 \times 4 + 12 \times \left( { - 6} \right)} \right)\\ \Rightarrow {S_{13}} = \frac{{13}}{2}\left( {8 - 72} \right)\\ \Rightarrow {S_{13}} = \frac{{13}}{2} \times \left( { - 64} \right)\\ \Rightarrow {S_{13}} = - 416\end{array}\]$

Hence the sum of first 13 term is $\[ - 416\]$.