Question

rupees 33000 speed of an 12 installments is that each installment is hundred more than that of the preceding 1 find the amount of the first and last installment​

Answer 1

It is a question of AP

So:

Sn =33000

n =12

d=100

a=?

l= ?

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

33000=12/2 [2a+(12-1)d]

33000= 6 [2a+11(100)]

33000/6= 2a +1100

5500-1100= 2a

4400 /2 = a

2200=a

l. = a +(n-1)d

=2200 +(11)(100)

=2200 +1100

=3300

Answer 2

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