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3. एक महिलाले एक महिनामा Rs. 32000 बचत गर्छिन्, अर्को महिनामा Rs. 36000 बचत गर्छिन् र तेस्रो महिनामा Rs. 40000 बचत गर्छिन्। यदि उनले यो क्रम जारी राख्छिन् भने,

a. यो अनुक्रमको समान अन्तर पत्ता लगाउनुहोस्।
[Ans: 4000]

b. चौथो महिनाको बचत पत्ता लगाउनुहोस्।
[Ans: 44000]

c. 3 वर्षको जम्मा बचत पत्ता लगाउनुहोस्।
[Ans: 36,72,000]

d. उनले कति महिनामा Rs. 20,00,000 बचत गर्लिन्?
[Ans: 20]

4. बाबुर दोराको उमेर For Q.N.g.
(Note: The context for this sentence is unclear. Please provide additional information for accurate translation and formatting.)

Sagot :

GinnyAnsweridn
Sure, let's go step-by-step to solve this question.

### Given Data:
- Savings in the 1st month: Rs. 32000
- Savings in the 2nd month: Rs. 36000
- Savings in the 3rd month: Rs. 40000

The savings form an arithmetic sequence where:
- [tex]\( a_1 = 32000 \)[/tex]
- [tex]\( a_2 = 36000 \)[/tex]
- [tex]\( a_3 = 40000 \)[/tex]

### a. Find the common difference of the sequence:

The common difference [tex]\( d \)[/tex] in an arithmetic sequence is given by:

[tex]\( d = a_2 - a_1 \)[/tex]

So,

[tex]\( d = 36000 - 32000 = 4000 \)[/tex]

Answer: [tex]\( d = 4000 \)[/tex]

### b. Find the saving in the 4th month:

The savings in the 4th month [tex]\( a_4 \)[/tex] can be found using the formula of the arithmetic sequence:

[tex]\( a_4 = a_3 + d \)[/tex]

So,

[tex]\( a_4 = 40000 + 4000 = 44000 \)[/tex]

Answer: [tex]\( a_4 = 44000 \)[/tex]

### c. Find the total saving in 3 years:

The total number of months in 3 years is:

[tex]\( 3 \times 12 = 36 \)[/tex] months

The sum [tex]\( S_n \)[/tex] of the first [tex]\( n \)[/tex] terms of an arithmetic sequence is given by:

[tex]\[ S_n = \frac{n}{2} [2a + (n-1)d] \][/tex]

For [tex]\( n = 36 \)[/tex],

[tex]\[ S_{36} = \frac{36}{2} [2 \times 32000 + (36-1) \times 4000] \][/tex]

[tex]\[ S_{36} = 18 [64000 + 140000] \][/tex]

[tex]\[ S_{36} = 18 \times 204000 \][/tex]

So,

[tex]\[ S_{36} = 3672000 \][/tex]

Answer: Rs. 3672000

### d. In how many months will she save Rs. 2000000?

We need to find the number of months [tex]\( n \)[/tex] it will take to save Rs. 2000000.

Using the sum formula for an arithmetic sequence:

[tex]\[ \frac{n}{2} [2a + (n-1)d] = 2000000 \][/tex]

[tex]\[ \frac{n}{2} [2 \times 32000 + (n-1) \times 4000] = 2000000 \][/tex]

[tex]\[ n(64000 + 4000(n-1)) = 4000000 \][/tex]

[tex]\[ n(64000 + 4000n - 4000) = 4000000 \][/tex]

[tex]\[ n(60000 + 4000n) = 4000000 \][/tex]

This simplifies to:

[tex]\[ 4000n^2 + 60000n - 4000000 = 0 \][/tex]

[tex]\[ n^2 + 15n - 1000 = 0 \][/tex]

Solving the quadratic equation, we get [tex]\( n = 25 \)[/tex], as it is the positive integer solution.

Answer: 25 months
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