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A person deposits rs. 1,000 in the first month. then every month he increases the monthly deposit by rs. 60. use the principle of progression and calculate

his total investment at the end of two years. find its solution


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
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Here the series is 1000,1060,1120,.............24 times
so here a=1000
   and n=24
then total investment is 
then =24/2[2*1000+(24-1)60]
        =40560 answer
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