Question

40th term of an ap whose 9th term is 465 and 20th term is 388

Answer 1

Term,9=465
Term, 20=388
Term 40= to find

a+8d=465.........................1
a+19d=388........................2

substracting both

11d=-77
d=-7

putting value of d in eqn. 1

a+8*-7=465
a-56=521
a=521


therefor
a=521

d=7


term 40 of an ap is

a+39d=521+39*-7

term 40=521-273

term 40=248



therefor term 40 of ap is 248.





Hope it helps
#/AJ#/