# A man saves Rs 50,000 during the year 2010 and at the end of 2010 he deposits it in a bank at 10% compound interest, compounded yearly. futher his savings increase by Rs 5000 every year which he deposits in the bank at the end of every year. find his total savings by the end of 2012

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by originallykool

2015-08-03T15:42:34+05:30

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Saving in 2010 = Amount deposited at the end of 2010 :  Rs 50, 000.
at 10% interest.compounded yearly
Amount accumulated at the end of 2012: Rs 50, 000 (1 + 10/100)² = Rs 60, 500

savings in 2011 = Amount deposited at the end of 2011 :  Rs 55, 000.
at 10% interest compounded yearly.
Amount accumulated at the end of 2012: Rs 55, 000 (1 + 10/100)¹ = Rs 60, 500.

Amount deposited at the end of 2012 :  Rs 60, 000. = savings during 2012..

Total amount of savings = Rs 60, 500 + Rs 60, 500 + Rs 60, 000
= Rs 181, 000

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If the savings and deposits are done like the above for n years, than the total savings at the end of n years could be found as below:

P = Rs 50, 000  ,      p =   Rs 5, 000    ,      r = 10% = 0.1                   1 + r = 1.1

S = P * 1.1^n + (P+p) * 1.1^{n-1} + (P+2 p) *1.1^{n-2} + ..... + [P + p(n-1)]* 1.1^{1} + (P + np)
= P * [1.1^n + 1.1^{n-1} + 1.1^{n-2} + .... + 1.1]
+ p [1.1^{n-1} + 2 * 1.1^{n-2} + 3 * 1.1^{n-3} + ... + (n-1) * 1.1 + n]

we apply the geometric progression formula here.

S = 11 * P [1.1^{n+1} - 1] + p [1.1^{n-1} + 2 * 1.1^{n-2} + 3 * 1.1^{n-3} + ... + (n-1) * 1.1 + n]
= 11 * P [1.1^{n+1} - 1]
+ (p * 1.1^{n}) * [1/1.1 + 2 / 1.1^2 + 3 / 1.1^3 + ... + (n-1) / 1.1^{n-1} + n / 1.1^{n} ]
= 11 * P [1.1^{n+1} - 1]
+ (p * 1.1^{n} * [

S = 10 P (1.1^{n+1} - 1] + 100 p [ 1.1^{n+1} - (n+1) 1.1 + n ]
= 10 * 50, 000 [1.1^3 - 1] + 100 * 5,000 [ 1.1^3 - 3 * 1.1 + 2]
= Rs 1, 81,000
this is the generic formula for recurring deposit P annually, with deposit amount increasing every year by amount p.