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. His total saving from the start of service will be Rs. 11040 after

Question

SHAIK AASIF AHAMED · Answer

Hello student, So after 21 months his total saving be Rs.11040 Thanks and Regards Shaik Aasif askIITians faculty

Rishi Sharma · Answer

Dear Student,
Please find below the solution to your problem.

Savings are:
200,200,200,240,280,.......uptonterms
But 200,240,280,......(n−2) terms forms an A.P
Since, sum of n terms of an AP =Sn=2n[2a+(n−1)d], where first term is a and common difference d
∴ According to question
400+2n−2[2(200)+(n−2−1)40]=11040 ......Sum =11040
⇒400+(n−2)[200+(n−3)20]=11040

⇒(n−2)[200+20n−60]=10640
⇒(n−2)(20n+140)=10640
⇒20(n+7)(n−2)=10640
⇒n2+5n−14=532
⇒n2+5n−546=0
⇒(n+26)(n−21)=0
⇒n=21
So, total time =21 months.

Thanks and Regards

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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. His total saving from the start of service will be Rs. 11040 after

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. His total saving from the start of service will be Rs. 11040 after

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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. His total saving from the start of service will be Rs. 11040 after

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. His total saving from the start of service will be Rs. 11040 after

2 Answers

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