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1-4/7 what is the answer |
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Answer» -STEP explanation:Here,givena=1d=4−1=3and,s n =287Now,s n = 2N (2a+(n−1)d)⇒287= 2n (2×1+(n−1)3)⇒287= 2n (2+3n−3)⇒574=n(3n−1)⇒574=3n 2 −n⇒3n 2 −n−574=0onsolvingthequadraticequatonusingformulan= 2a−b± b 2 −4ac Wegetn=14& 3−41 [doesnotexist]so,n=14Now,s n = 2n (a+1)⇒287= 214 (1+x)⇒574=14(1+x)⇒(1+x)= 14574 ⇒1+x=41⇒x=41−1∴x=40x=40isthesolution. |
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