Question

33000 paid of in 12installment, such that each installment is Rs 100 more than the preceding find the amount of first and the last installment

Answer 1

Answer:

First installment = 2200

Last installment = 3300

Step-by-step explanation:

Total paid: 33,000

Let first installment be r

2nd will be r+100

3rd will be r+200

12th will be r+1100

So,

12r + (100 + 200 + .. + 1100) = 33000

12r + 11(11+1)/2 = 33000

12r + 6600 = 26400

r = 26400 / 12 = 2200

So,

First installment = 2200

Last installment = 2200 + 1100 = 3300

Answer 2

Step-by-step explanation:

let the first and the last installments be a and l resp.

given,

Sn=33000, n=12, d=100

a=?, l=?

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

33000=12/2{2a+(12-1) 100}

33000=6{2a+1300}

33000/6=2a+1300

5500-1300=2a

4200/2=a

a=2100

now

Sn=n/2(a+l)

33000=12/2(2100+l)

33000/6=2100+l

5500-2100=l

l=3400

so the first and the last terms are 2100 and 3400 resp.

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