Question

Solve that sum if you Can..Good Explanation will be mark as brainliest...

Answer 1

hey there,
If I'm understanding correctly, the question states that the interest paid annually is 12% of 6000, 12% of 5500, 12% of 5000, ..., 12% of 500. From this we infer that either the interest payable each year is calculated before the Rs 500500 for that year is subtracted from the amount left to pay, or that annual payments of Rs 500500 don't start until the second year. Either way, the answer is the same.The farmer pays Rs 6000 to begin with, leaving Rs 6000 to pay, so the interest payable in the first year is 12% of 6000. In year 2, he pays Rs 500, leaving Rs 6000−500=6000−500= Rs 55005500 to pay, so the interest for year 2 is 12% of 5500. In year 3, he pays Rs 500, leaving Rs 6000−500(2)=6000−500(2)= Rs 50005000 to pay, so the interest for year 3 is 12% of 5000. ,..., In year 12, he pays Rs 500, leaving 6000−500(11)=6000−500(11)= Rs 500500 to pay, so the interest for year 12 is 12% of 500. The interest payments form an arithmetic progression with a1=600012100=720a1=600012100=720 and d=−50012100=−60d=−50012100=−60 , so the total interest payments are s12=122(720+60)=s12=122(720+60)= Rs Hope this helps!

Answer 2

Hello buddy..

Answer :

Amount Paid to buy tractor = Rs. 12,000

Farmer Pays Cash = Rs. 6000

Remaining Balance = 12000 - 6000 = 6000

Annual Instalment = Rs 500 + interest@12% on unpaid amount

1st Instalment

Unpaid Amount = Rs. 6000

Interest on Unpaid Amount = (12/100) × 6000 = 720

Amount of Instalment = Rs. 500 + Rs. 720 = Rs. 1220

2nd Instalment

Unpaid Amount = Rs. (6000 - 500) = Rs. 5500

Interest on Unpaid Amount = (12/100) × 5500 = 6600

Amount of Instalment = Rs. 500 + Rs. 660 = Rs. 1160

3rd Instalment

Unpaid Amount = Rs. (5500 - 500) = Rs. 5000

Interest on Unpaid Amount = (12/100) × 5000 = 600

Amount of Instalment = Rs. 500 + Rs. 600 = Rs. 1100

Total no. of Instalments = 6000/500 = 12

Thus, Annual Instalments are 1220, 1160, 1100, …upto 12 terms

Since the common difference between the consecutive terms is constant. Thus, Annual Instalments are in AP.

Here

first term(a) = 1220

Common difference(d) = 1160 - 1220 = - 60

Number of terms(n) = 12

Total amount paid in 12 instalments is given by -

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

∴ S12 = (12/2)[2(1220) + (12 - 1)( - 60)]

= 6[2440 + 11( - 60)]

= 6[2440 - 660]

= 6 × 1780

= 10680

Hence, total amount paid in 12 Instalments = Rs 10680

Hence,

Total Cost of Tractor

= Amount paid earlier + Amount paid in 12 Instalments

= Rs. (6000 + 10680)

= Rs. 16680

Hope this helps u dude ⭐

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