Maturity value of recurring deposits - class-X

Let the monthly installment be $P$

Using the formula,
$M =\dfrac{R \times [(1+i)^{n} - 1]}{1-(1+i)^{\dfrac{-1}{3}}}$ 
R $=$ Monthly Installment $= 2000 $
i $=$ Rate of Interest $5%$
n $=$ Number of Quarters $= 1$
Substitute the values, 
$M =\dfrac{2000 \times [(1+5/400)^{n} - 1]}{1-(1+5/400)^{\frac{-1}{3}}}$ 
$M = \dfrac{2000\times\dfrac{5}{400}}{1-(\dfrac{405}{400})^{-1/3}}$
$M = 6049.931$

Using the formula,
$M =\dfrac{R \times [(1+i)^{n} - 1]}{1-(1+i)^{\frac{-1}{3}}}$ 
$R =$ Monthly Installment $= 200$ 
$i =$ Rate of Interest/ 400 $= \dfrac {2}{400}$
$n =$ Number of Quarters $= 1$
Substitute the values, 
$M =\dfrac{200 \times [(1+2/400)^{n} - 1]}{1-(1+2/400)^{\frac{-1}{3}}}$ 
$M = \dfrac{200\times\dfrac{2}{400}}{1-(\dfrac{402}{400})^{-1/3}}$
$M = 601.99$

Recurring deposit type of account is operated by salaried persons and petty traders.

Using the formula,
$M =\dfrac{R \times [(1+i)^{n} - 1]}{1-(1+i)^{\frac{-1}{3}}}$ 
$R $= Monthly Installment $= 1000 $
$i =$ Rate of Interest/ 400 $= \dfrac {4}{400}$
$n =$ Number of Quarters $= 1$
Substitute the values, 
$M =\dfrac{1000 \times [(1+4/400)^{n} - 1]}{1-(1+4/400)^{\frac{-1}{3}}}$ 
$M = \dfrac{1000\times\dfrac{4}{400}}{1-(\dfrac{404}{400})^{-1/3}}$
$M = 3019.97$

Given, maturity value $=146437.5$, monthly installment $=5,500$, $n=2$ years $=2\times 12=24$ months

Given: $P = 100, T(n) = 12$ months, $R = 10%$
Equivalent principal for $12$ months $=$ $100\times \dfrac{n(n+1)}{2}$
$=$ $100\times\dfrac{12(13)}{2}$
$= 50 \times  12 \times  13$
Interest $=$ $\dfrac{PRT}{100}$
$=$ $\dfrac{50\times 12 \times 13\times 10\times 1}{100\times 12}$
$=$ $65$
Maturity amount $=$ P$\times$T $+$ Interest
$=$ $100 \times 12 + 65$
$=$ Rs. $1265$

Let the monthly installment be $P$

Using the formula,
$M =\dfrac{R \times [(1+i)^{n} - 1]}{1-(1+i)^{\frac{-1}{3}}}$ 
$R =$ Monthly Installment $= 700 $
$i =$ Rate of Interest/ 400 $= \dfrac {10}{400}$
$n =$ Number of Quarters $= 1$
Substitute the values, 
$M =\dfrac{700 \times [(1+10/400)^{n} - 1]}{1-(1+12/400)^{\frac{-1}{3}}}$ 
$M = \dfrac{700\times\dfrac{10}{400}}{1-(\dfrac{410}{400})^{-1/3}}$
$M = 2134.904$

Using the formula,
$M =\dfrac{R \times [(1+i)^{n} - 1]}{1-(1+i)^{\frac{-1}{3}}}$ 
$R =$ Monthly Installment $= 400$ 
$i =$ Rate of Interest/ 400 $= \dfrac {12}{400}$
$n =$ Number of Quarters $= 1$
Substitute the values, 
$M =\dfrac{400 \times [(1+12/400)^{n} - 1]}{1-(1+12/400)^{\frac{-1}{3}}}$ 
$M = \dfrac{400\times\dfrac{12}{400}}{1-(\dfrac{412}{400})^{-1/3}}$
$M = 1223.92$

Given, $P =$ Rs. $200$, $n = 6 $ months, $r = 7%$
Interest $=$ $\dfrac{Pn(n+1)r}{2400}$

Let the rate of interest per annum be $r%$.
$P =$ Rs. $2500, n = 48$ months
Interest $=$ $\dfrac{Pn(n+1)r}{2400}=\dfrac{2500\times48(48+1)r}{2400}= 2540r$
Maturity amount $ = 2500 \times 48 + 2450r = 120000 + 2450r$
$\Rightarrow 150000 - 120000 = 2450r$
$\Rightarrow 30000 = 2450r$
$\Rightarrow r = 12.24%$

In recurring deposit account, a depositor chooses a fixed amount and deposits it for fixed period every month.

$A=P(1+\cfrac{r}{100})^n$

It is given that

$Let\quad the\quad initial\quad deposit\quad be\quad Rs.\quad x,\ amount\quad after\quad 1st\quad year=Rs.\quad 2x\ amount\quad left\quad after\quad withdrawl=Rs.\quad 2x-8000,\ amount\quad after\quad 2nd\quad year=Rs.\quad 4x-16000\ amonut\quad left\quad after\quad withdrawl=Rs.\quad 4x-24000,\ amount\quad after\quad 3rd\quad year=Rs.\quad 8x-48000=Rs.\quad 8000,\ 8x=56000,\ x=Rs.\quad 7000$

The fixed period of recurring deposit  account can be $6$ months or more.

Let the monthly installment be $P$.