Question

Please solve question 29 on applications of AP ( Arithmetic Progression ) ​

Answer 1

Answer:

a = 450 (first instalment)

d = 50

no. of installments is 15 so last installment will be

an = a + (n-1)d

= 450 + (15-1) 50

= 450 + 14×50

= 450 + 700

= 1150

Her last instalment was ₹1150

Answer 2

Answer:

Last installment - Rs.1150

Total Amount Paid - 12000 Rs.

Step-by-step explanation:

In AP each term can be written as

a1 + (n-1) d

where a1 is first term of AP

n is the number of the term which is to be written

d is the common difference of AP

Sum of AP = N/2(a1+aN)

here N is the total number of terms in AP

aN is the last term of AP

In this question,

N = 15

a1= 450

d= 50

to find-

aN last term

Sum of AP

Solution- aN = a1 + (N-1)d

= 450 + 14*50

= 450 + 700

aN= 1150

Sum = N/2(a1+aN)

= 15/2(450+1150)

=15/2(1600)

= 15*800

= 12000