A man borrows Rs. 4,500 and promises to pay back in 30 installments, each of value Rs. 10 more than the last. Find the 1st and last installments.

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A man borrows Rs. 4,500 and promises to pay back in 30 installments, each of value Rs. 10 more than the last. Find the 1st and last installments.
    
Solution:
Here the summation amount is 4500. S=4500.
d=10
n=30.
We assume that first installment=a.
We will use the summation formula.
    S=(n÷2)×{2a+(n-1)×d}
=>4500=(30÷2)×{2a+(30-1)×10}
=>4500=15×{2a+29×10}
=>4500=15×{2a+290}
=>4500÷15=2a+290
=>300=2a+290
=>300-290=2a
=>10=2a
=>10÷2=a
=>5=a
So the first installment=Rs. 5

The last installment is 
t₃₀=a+(30-1)×d
        =5+29×10
        =5+290
        =295.


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