Question

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Answer 1

Question :- Neha and Swati made an agreement. Neha would give ₹100 to Swati everyday. Swati would start with ₹1 and give Neha double the amount of the previous day. After 10 days, who will have more money? How much more?

Solution :-

Given, Neha would give ₹100 everyday. So after 10 Days, she would have given ₹100 × 10 = ₹1000 to Swati.

Now Swati gives double of the previous. She started with 1.

So, first day Swati gave Neha rupee 1 = $2^{0}$

Second day 1 × 2 = 2 = $2^{1}$

Third day 2 × 2 = 4 = $2^{2}$

and so on.

Common ratio =
$\frac{T_{2}}{T_{1}} = \frac{T_{3}}{T_{4}}$
=
$\frac{2}{1} = \frac{4}{2} \\ \\ = > 2 = 2$

Since the common ratio is equal, the given progression is G.P.

The G.P. is

1, 2, 2^2, 2^3....

Sum of G.P if common ratio > 1

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

Here, n = 10
a = 1
r = 2

=> Sum =
$\frac{1( {2}^{10} - 1) }{(2 - 1)} \\ \\ = {2}^{10} - 1 \\ \\ = 1024 - 1 \\ \\ = 1023$

So, Swati gave Nidhi ₹1023 after 10 Days, and Nidhi gave ₹1000.
Answer :- So, Swati gave ₹23 more to Nidhi.


Method 2



Nidhi gives ₹100 everyday. So after 10 days, she would give ₹1000.

Now, for first day Swati gives ₹1

Second day she gives ₹1 × 2 = ₹2

Third Day = ₹2 × 2 = ₹4

Fourth day = ₹4 × 2 = ₹8

Fifth Day = ₹8 × 2 = ₹16

Sixth day = ₹16 × 2 = ₹32

Seventh day = ₹32 × 2 = ₹64

Eighth day = ₹64 × 2 = ₹128

Ninth day = ₹128 × 2 = ₹256

Tenth Day = ₹256 × 2 = ₹512

So, sum = 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512

= ₹1023

So Swati gave ₹1023 after 10 days.

How much more she give?

=> 1023 - 1000 = ₹23

Answer :- Swati gives more rupees by 23.

Answer 2

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