Question

If the 20^th term from the end of the progression 20,19\({\frac{1}{4}}\) ,18\({\frac{1}{2}}\) ,17\({\frac{3}{4}}\) ,……, -129\({\frac{1}{4}}\) is______?

• A. -120
• B. -115
• C. -125
• D. -110

Answer 1

The Correct Option is B

A.P.  20,19\({\frac{1}{4}}\) ,18\({\frac{1}{2}}\) ,17\({\frac{3}{4}}\) ,……, -129\({\frac{1}{4}}\) 

This is A.P. with common difference 
\(d_1 = 19\frac1{4}-20 = − \frac3{4 }\)

⇒ This is also A.P. -129\({\frac{1}{4}}\) ,…………,19\({\frac{1}{4}}\),20

So, a=-129\({\frac{1}{4}}\)  and d =\(\frac3{4 }\)

20^th term = -129\({\frac{1}{4}}\) + (20-1)( \(\frac3{4 }\) )

=\(-\frac{517}4 + 15-\frac3{4}\)

=-130+15

=-115

Hence, The correct option is (B) : -115