Question

depoist=Rs 1000p/m in 3years at 8% p.a, Find maturity value * ​

Answer 1

Given :

TheAmount deposited per month = p = Rs 1000

The rate of interest applied = r = 7 %

The maturity value = Rs 70675

The number of months = n

To Find :

The value of n

Solution :

According to question

∵ Interest = principal × \dfrac{n(n+1)}{2}2n(n+1) × \dfrac{RT}{100}100RT

where n = number of months

          R = rate of interest

          T = \dfrac{1}{12}121

So, Interest = Rs 1000 × \dfrac{n(n+1)}{2}2n(n+1) ×  \dfrac{7\times 1}{100\times 12}100×127×1

Or,  Interest = Rs 1000 × \dfrac{n(n+1)}{2}2n(n+1) × \dfrac{7}{1200}12007

Or,  Interest =  \dfrac{n(n+1)}{2}2n(n+1) × \dfrac{70}{12}1270

∴   Interest =  {n(n+1)}n(n+1) × \dfrac{35}{12}1235

Again

Maturity value = Interest + monthly deposit money × number of month

70675    = {n(n+1)}n(n+1) × \dfrac{35}{12}1235 + Rs 1000 × n

Or,  70675 × 12 = 35 n ( n + 1 ) + 1000 n  × 12

or,  848100 = 35 n² + 35 n + 12000 n

Or,  35 n² + 12035 n - 848100 = 0

Or,    7 n² + 2407 n - 169620 = 0

Solving this quadratic equation

  n = \dfrac{-2407\pm \sqrt{2407^{2}-4\times 7\times (-169620))}}{2\times 7}

2×7−2407±24072−4×7×(−169620))

So, n = 60  ,  - 403

So, Number of months = n =  60

Hence, The value of n is 60  Answer