Exams Joint Entrance Examination Mathematics Sequences and Series Question 70 Question 70 of 519 Easy MCQ
20 20 , 19 1 4 , 18 1 2 , 17 3 4 , … , − 129 1 4 {20}^{}20,19\frac{1}{4},18\frac{1}{2},17\frac{3}{4},\ldots ,-129\frac{1}{4} 20 20 , 19 4 1 , 18 2 1 , 17 4 3 , … , − 129 4 1
Solutions · 1 Post your solution Official solution
20 , 19 1 4 , 18 1 2 , 17 3 4 , … … , − 129 1 4 20,19\frac{1}{4},18\frac{1}{2},17\frac{3}{4},\ldots \ldots ,-129\frac{1}{4} 20 , 19 4 1 , 18 2 1 , 17 4 3 , …… , − 129 4 1
This is A.P. with common difference
d 1 = − 1 + 1 4 = − 3 4 − 129 1 4 , … … … … . . .19 1 4 , 20 \begin{gathered} {d}_{1}=-1+\frac{1}{4}=-\frac{3}{4} \\ -129\frac{1}{4},\ldots \ldots \ldots \ldots ...19\frac{1}{4},20 \end{gathered} d 1 = − 1 + 4 1 = − 4 3 − 129 4 1 , ………… ...19 4 1 , 20
This is also A.P. a = − 129 1 4 a=-129\frac{1}{4} a = − 129 4 1 and d = 3 4 d=\frac{3}{4} d = 4 3
Required term = = =
− 129 1 4 + ( 20 − 1 ) ( 3 4 ) = − 129 − 1 4 + 15 − 3 4 = − 115 \begin{gathered} -129\frac{1}{4}+(20-1)(\frac{3}{4}) \\ =-129-\frac{1}{4}+15-\frac{3}{4}=-115 \end{gathered} − 129 4 1 + ( 20 − 1 ) ( 4 3 ) = − 129 − 4 1 + 15 − 4 3 = − 115 Answer the question to unlock