Arithmetic Sequence Need Assistance

asked Oct 16, 2020 11.0k views

1 vote

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6 votes

let the first term is

by the sum of first n term formula

$(20)/(2)(2a+19d)=25a\\20a+190d=25a15\\190d=5a\\a=38d$

substitute to sum of first 30 term

$(30)/(2)(2a+29d)=(30)/(2)(2(38d)+29d)\\=15(76d+29d)\\=15(105d)\\=1575d$

Umair Durrani answered Oct 22, 2020

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