Question

a man promises to repay Rs 3,27,600 in 12 monthly instalments. the value of each instalment is double the earlier one . calculate the 1st instalment and the last one

Answer 1

Hello dude,

Here is your answer,

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Let the installments be in a G.P(Geometric Progression)

Let 'a' be the first term

'r'be the common ratio

$S_{n}$ = $\frac{a(r^{n}-1 )}{r-1}$

We  know that r = 2

, n=12 , Sum =₹3,27,600

Putting the values:

327600 = $\frac{a(2^{12} -1)}{2-1}$

327600 = $a(4096 - 1)$

a = $\frac{327600}{4095}$

a = ₹80

Hence the first Installment is 80.

Let the Last Installment be $t_{n}$

$t_{n}$ = $ar^{n-1}$

Putting the values

$t_{n}$ = $80$×$2^{11}$

$t_{n}$ =  $80$×$2048$

$t_{n}$ =₹ $163840$

Thus the  last installment is₹ 163840

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