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} * [