Answers

2014-03-31T22:35:58+05:30
Given that,
a=first term=1000
r=increase in the monthly deposit=60
we know for an A.P sequence, sum upto n terms=n[a+(n-1)d]/2
so as we have 1 year= 12 months => 2 yrs = 24 months,
  we get n=24
hence total investment at the end of 2 years is given by,
   =>     n[a+(n-1)d]/2
   =>     24[1000+23*60]/2
   =>     24[2380]/2
   =>     12*2380
   =>     28560
0
2014-04-01T11:37:47+05:30
Here the series is 1000,1060,1120,.............24 times
so here a=1000
  d=60
   and n=24
then total investment is 
Sn=N/2[2a+(n-1)d]
then =24/2[2*1000+(24-1)60]
      =12(2000+1380)
       =3380*12
        =40560 answer
0