Question

₹33,000 is paid off in 12 instalments, such that
each instalment is 100 more than the preceding
one. Find the amounts of the first and the last
instalments.​

Answer 1

Step-by-step explanation:

they are in A.P

Sn = 33000, n = 12, d = 100, a = ?, a12 = ?

Sn = n/2 [ 2a + (n-1) d ]

33000 = 12/2 [ 2a + 11×100]

33000 = 6 [ 2a + 1100]

33000/6 = 2a + 1100

5500 = 2a + 1100

5500 - 1100 = 2a

4400 = 2a

therefore a = first installment = Rs 2200

A12 = last installment

A12 = a + (n - 1) d

A12 = 2200 + 11×100

A12 = 2200 + 1100 = Rs 3300

Answer 2

Step-by-step explanation:

In terms of A.P:

Given: An=33000, n=12, d=100.

To find: a=? l=?

Formula: Sn=n/2(2a+(n-1)d)

An=a+(n-1)d.

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

33000/6=2a+1100

5500-1100=2a

4400/2=2200=a.

A12=2200+(12-1)100

=2200+1100=3300.

Amount of first installment=2200.Rs

Amount of last installment=3300.Rs