Question

Which term of a.p 5,10,15....exceeds its 31 st term by 130

Answer 1

here,

a=5

d=  10-5 = 5

Therefore,

a 31  =  a + (n-1)d

        = 5 + 30* 5

        =5 + 150

        =155

now,

the term which exceeds it's 31 term=   a + (n-1)d

since,     a n= 155 + 130 = 285

now ,

an = a+(n-1)d

285=5+(n-1)5

280=(n-1)5

280/5 = n-1

56 = n-1

n = 56+1

n = 57  

hence ,

57 term of ap exceeds it's  31 term by 130