Question

a man saves rs.200 in each of the 1st 3 months of his service. in each of the subsequent months his saving increases by rs.40 more than the saving of immediately previous month.his total saving from the start of the service will rs.11040 after how many months

Answer 1

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☞ Your Answer is 21 months

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$\huge\sf\blue{Given}$

✭ A man saves Rs 200 in each of the first 3 months

✭ In the remaining months he save Rs 40

✭ Total savings = 11040

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$\huge\sf\gray{To \:Find}$

➢ Number of months?

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So here it is an AP, So,

➢ A = 200

➢ D = 40

Saving in the first 2 months = Rs 400

So then the remaining savings is,

➳ 200,240,280,320.... upto the nth term

So now the sum,

➝ $\sf \dfrac{n}{2}\bigg\lgroup 2×200 +(n-1)40 = 11040 - 400\bigg\rgroup$

➝ $\sf 200n+20n^2 -20n = 10640$

➝ $\sf 20n^2 - 180n - 10640 = 0$

➝ $\sf n^2 + 5n - 546 = 0$

➝ $\red{\sf (n-21)(n+26) = 0}$

➝ $\sf \orange {n = 21} \qquad\bigg\lgroup Ignoring \ negative \ sign \bigg\rgroup$

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Answer 2

In the above Question , the following information is given -

A man saves Rs. 200 in each of the first three months of his service .

In each of the subsequent months his saving increases by Rs.40 more than the saving of immediately previous month.

To find -

After how many months his total savings from the start of service will be Rs.11040 ?

Solution -

Here ,

A man saves Rs. 200 in each of the first three months of his service .

In each of the subsequent months his saving increases by Rs.40 more than the saving of immediately previous month.

So,

For the first three months -

Savings -

200 , 200 , 200

After that -

240 , 280 , 320 ,. ...

Sum of salary for the first two months -

=> 400 .

Remaining Savings -

=> 11040 - 400

=> 10640 .

Now ,

This series can be written as an Ap

Here ,

a = 200

d = 40

Sum till n terms = 10640

Now ,

Sum of n terms of an ap -

=> { n / 2 } ( 2a + ( n - 1 ) d } = 10640

Substituting the given Values -

=> ( n - 2 )( 20 n + 140 ) = 10640

=> 20 ( n - 2 )( n + 7 ) = 10640

=> ( n - 2 )( n + 7 ) = 532

=> n² + 5 n - 14 = 532

=> n² + 5n - 546 = 0

=> n² + 26n - 21n - 546 = 0

=> n ( n + 26 ) - 21 ( n + 26 ) = 0

=> ( n - 21 )( n + 26 ) = 0

Now , .

As n can't be negative , n = 21.

Thus , we can say that after 21 months , he will save the required amount .

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