Given-that-18-24-and-k-14-are-three-consecutive-terms-of-an-arithmetic-progression-Find-a-the-common-difference-b-the-value-of-k-c-the-first-term-if-the-4-th-term-is-12-d-the-sum-of-the-fir

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Question Number 35594 by Rio Mike last updated on 20/May/18

Given that 18,24,and k + 14 are three consecutive terms of an arithmetic progression Find a) the common difference b) the value of k c)the first term if the 4^(th) term is 12. d) the sum of the first twelve terms of the progression.

$${Given}\:{that}\:\mathrm{18},\mathrm{24},{and}\:{k}\:+\:\mathrm{14}\:{are}\: \\ $$$${three}\:{consecutive}\:{terms}\:{of}\:{an}\: \\ $$$${arithmetic}\:{progression}\:{Find} \\ $$$$\left.{a}\right)\:{the}\:{common}\:{difference} \\ $$$$\left.{b}\right)\:{the}\:{value}\:{of}\:{k} \\ $$$$\left.{c}\right){the}\:{first}\:{term}\:{if}\:{the}\:\mathrm{4}^{{th}} \:{term}\:{is} \\ $$$$\mathrm{12}. \\ $$$$\left.{d}\right)\:{the}\:{sum}\:{of}\:{the}\:{first}\:{twelve}\:{terms}\:{of} \\ $$$${the}\:{progression}. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 20/May/18

a)c.d=6 b)k+14=30 k=16 c)a+(4−1)×6=12 a=−6 d)s=((12)/2){2×−6+(12−1)×6} =6(−12+66) =6×54 =324

$$\left.{a}\right){c}.{d}=\mathrm{6} \\ $$$$\left.{b}\right){k}+\mathrm{14}=\mathrm{30} \\ $$$${k}=\mathrm{16} \\ $$$$\left.{c}\right){a}+\left(\mathrm{4}−\mathrm{1}\right)×\mathrm{6}=\mathrm{12} \\ $$$${a}=−\mathrm{6} \\ $$$$\left.{d}\right){s}=\frac{\mathrm{12}}{\mathrm{2}}\left\{\mathrm{2}×−\mathrm{6}+\left(\mathrm{12}−\mathrm{1}\right)×\mathrm{6}\right\} \\ $$$$=\mathrm{6}\left(−\mathrm{12}+\mathrm{66}\right) \\ $$$$=\mathrm{6}×\mathrm{54} \\ $$$$=\mathrm{324} \\ $$