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In-an-A-P-the-sum-of-the-first-50-terms-is-6275-Write-this-A-P-knowing-that-the-ratio-is-5-




Question Number 55628 by Mikael_Marshall last updated on 28/Feb/19
In an A.P, the sum of the first 50 terms is 6275. Write  this A.P . knowing that the ratio is 5.
$${In}\:{an}\:{A}.{P},\:{the}\:{sum}\:{of}\:{the}\:{first}\:\mathrm{50}\:{terms}\:{is}\:\mathrm{6275}.\:{Write}\:\:{this}\:{A}.{P}\:.\:{knowing}\:{that}\:{the}\:{ratio}\:{is}\:\mathrm{5}. \\ $$
Answered by kaivan.ahmadi last updated on 28/Feb/19
s_n =(n/2)[2a+(n−1)d]  6275=25[2a+49×5]=25[2a+245]=  50a+6145⇒50a=6275−6145=150⇒  a=3  the sequence is  3,8,13,18,23,...  a_n =a+(n−1)d=3+(n−1)5⇒  a_n =5n−2
$${s}_{{n}} =\frac{{n}}{\mathrm{2}}\left[\mathrm{2}{a}+\left({n}−\mathrm{1}\right){d}\right] \\ $$$$\mathrm{6275}=\mathrm{25}\left[\mathrm{2}{a}+\mathrm{49}×\mathrm{5}\right]=\mathrm{25}\left[\mathrm{2}{a}+\mathrm{245}\right]= \\ $$$$\mathrm{50}{a}+\mathrm{6145}\Rightarrow\mathrm{50}{a}=\mathrm{6275}−\mathrm{6145}=\mathrm{150}\Rightarrow \\ $$$${a}=\mathrm{3} \\ $$$${the}\:{sequence}\:{is} \\ $$$$\mathrm{3},\mathrm{8},\mathrm{13},\mathrm{18},\mathrm{23},… \\ $$$${a}_{{n}} ={a}+\left({n}−\mathrm{1}\right){d}=\mathrm{3}+\left({n}−\mathrm{1}\right)\mathrm{5}\Rightarrow \\ $$$${a}_{{n}} =\mathrm{5}{n}−\mathrm{2} \\ $$
Answered by Mikael_Marshall last updated on 28/Feb/19
thank you Sir.
$${thank}\:{you}\:{Sir}. \\ $$

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