Question

❤️Please answer this question-

A farmer borrows Rs 4000 and agrees to repay with interest Rs 500 in 10 installments , each installment being less than the preceding one by Rs. 10 , then find the amount of 1st and last installments.❤️

No spam answers.

Answer 1

Hello...



Please refer to the attachment above...


Thank you ...

Answer 2

here is your answer OK ☺☺☺☺☺


Let the First Instalment be 'a'.
and second Instalment be '10-a'.
Since it will form an A.P.

The sum of ten Instalments = 10/2[2a + (10-1) (-10)] = 5[2a - 90]
So, 5 [2a-90] = 4000.
= 10a - 450 = 4000.
10a = 4450.
a = 4450/10
a = 445.
First Instalment will be Rs. 445.

second process..........

farmer borrow = Rs 4000Interest to be given by farmer = Rs 500Total money repaid by farmer= 4000 + 500 = Rs 4500Now, he repays this money in 10 equal instalmentsNow, number of instalments, n = 10Let the amount of first instalment = Rs xamount of second instalment = Rsx-10amount of third instalment = Rsx-20and so on.Now, the list of amount of instalments in Rs is :x, x-10, x - 20, ..........The above list forms an AP with first term, a = x and common difference, d = -10Now, we have, number of instalments, n = 10Now, sum of 10 instalments = Rs 4500Now, we know that sum of first n terms of AP isSn = n22a + n-1dSo, S10 = 4500⇒1022×x + 10-1-10 = 4500⇒52x-90 = 4500⇒2x - 90 = 900⇒2x-45 = 900⇒x - 45 = 450⇒x = 495So, amount of first instalment = Rs 495


i