Question

man borrows rupees 8000 and agrees to repay with a total interest of rupees 1360 in 12 monthly instalments. Each instalment being less than the preceding one by rupees 40. Find the amount of the first and last instalment. Solve the word problem

Answer 1

Answer:

First installment: 1000/-

Last installment: 560/-

Step-by-step explanation:

Amount borrows = 8000/-

Interest = 1360/-

Total amount = 9360/-

n= 12

Difference between two installments = 40/-

ie d= -40

$S_n = \frac{n}{2} \bigg(2a + (n - 1)d\bigg) \\ \\ 9360 = \frac{12}{2} \bigg(2a - (12 - 1)40\bigg) \\ \\ 9360 = 6(2a - 440) \\ \\ \frac{9360}{6} = 2a - 440 \\ \\ 1560 = 2a - 440 \\ \\ 2a = 2000 \\ \\ a = \frac{2000}{2} \\ \\ a = 1000$

First installment is 1000 Rs

Second installment = 1000-40= 960

last installment:12 th

general expression for nth term

$T_n= a+(n-1)d$

$T_{12} = 1000 + (11)(-40) \\ \\ = 1000 - 440 \\ \\ = 560$

Hope it helps you.