The 20 th term from the end of the progression 20 , 19 1 4 , 18 1 2 , 17 3 4 , … , - 129 1 4 is: [2024]

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Solution

Q.
The $20^{th}$ term from the end of the progression $20, 19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \dots, - 129 \frac{1}{4}$ is: [2024]

1 $- 118$

2 $- 100$

3 $- 110$

4 $- 115$

Ans.
(4)

Given, $20, 19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \dots, - 129 \frac{1}{4}$ is in A.P. with first term $(a) = - 129 \frac{1}{4}$

Common difference $(d) = 20 - 19 \frac{1}{4} = \frac{80 - 77}{4} = \frac{3}{4}$

$20^{th}$ term from the end $= a_{20} = a + (20 - 1) d$

$= - 129 \frac{1}{4} + 19 \times \frac{3}{4} = \frac{- 517}{4} + \frac{57}{4} = \frac{- 460}{4} = - 115$

$∴$ $a_{20} = - 115$

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