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

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