Question

Dajshi deposited rs 1000 per month for n months in a bank's rd account. if the bank pays the rate of interest 7% annually and the maturity value is rs 70675, find the value of n.

Answer 1

Given :

The Amount 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 × $\dfrac{n(n+1)}{2}$ × $\dfrac{RT}{100}$

where n = number of months

          R = rate of interest

          T = $\dfrac{1}{12}$

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

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

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

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

Again

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

70675    = ${n(n+1)}$ × $\dfrac{35}{12}$ + 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 = $\dfrac{-2407\pm \sqrt{2407^{2}-4\times 7\times (-169620))}}{2\times 7}$

So, n = 60  ,  - 403

So, Number of months = n =  60

Hence, The value of n is 60  Answer

Answer 2

Answer:

60

Step-by-step explanation:

Interest = principal ×  ×

where n = number of months

         R = rate of interest

         T =

So, Interest = Rs 1000 ×  ×  

Or,  Interest = Rs 1000 ×  ×

Or,  Interest =   ×

∴   Interest =   ×

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

70675    =  ×  + 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}

So, n = 60  ,  - 403

So, Number of months = n =  60

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