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Sequences and series

Anyone know how to calculate this question?

The sum of the first six terms of an arithmetical progression is 21, and seventh term is three times the sum of the third and fourth. Find the first term and the common difference.
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NDVA
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Original post by izzy610
Anyone know how to calculate this question?

The sum of the first six terms of an arithmetical progression is 21, and seventh term is three times the sum of the third and fourth. Find the first term and the common difference.


Let a1,a2,...,ana_1, a_2, ..., a_n be the first n terms of this series and dd be the common difference. Then

S6=21=12×6×(a1+a6)=12×6×(a1+a1+5d)S_6 = 21 = \frac{1}{2} \times 6 \times (a_1 + a_6) = \frac{1}{2} \times 6 \times (a_1 + a_1 + 5d)

⇒7=2a1+5d\Rightarrow 7=2a_1 + 5d


Also

a7=3(a3+a4) a_7 = 3 (a_3+a_4)

⇒a1+6d=3(a1+2d+a1+3d)\Rightarrow a_1 + 6d = 3(a_1 + 2d + a_1 + 3d)

⇒0=5a1+9d\Rightarrow 0 = 5a_1 + 9d


Hence we have the simultaneous equations

7=2a1+5d[br]0=5a1+9d7=2a_1 + 5d[br]0 = 5a_1 + 9d

It follows that a1=−9a_1=-9 and d=5d=5 :smile:

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