Jaspal singh repays his total loan of Rs, 118000 by paying every month starting with the first instalment of Rs.1000. If he increases the instalment by Rs. 100 every month , what amount will be paid by him in the 30th instalment ? What amount of loan des he still have to pay after the 30th instalment?
Answers
Answered by
472
Solution:-
Let us assume that the loan is cleared in 'n' months.
Then, The amounts are in AP with first term (a) = 1000, the common difference (d) = 100 and the Sum of amounts (Sn) = Rs. 118000
So, according to the question.
Sn = n/2{2a + (n - 1)d}
118000 = n/2{2 × 1000 + (n - 1)100}
236000 = n{2000 - 100 + 100n}
236000 = 1900n + 100n²
100n² + 1900n - 236000 = 0
Dividing the above by 100, we get.
n² + 19 - 2360 = 0
n² + 59n - 40n - 2360 = 0
n(n + 59) - 40(n +59) = 0
(n - 40) (n + 59) = 0
n = 40 or n = -59
Since the value of cannot be negative so, n = 40 is correct.
Therefore, the loan will be cleared in 40 months.
Now, the amount to be paid by him in the 30th installment is calculated as under.
a₃₀ = a + 29d where a = 1000 and d = 100
a₃₀ = 1000 + (29*100)
= 1000 + 2900
a₃₀ = 3900
so, the amount to be paid by him in the 30th installment is Rs. 3900
Now, the amount of remaining loan that he has to pay after the 30th installment is calculated as under.
S₃₀ = n/2{2a + (n - 1)d}
= 30/2{2 × 1000 + (30 - 1)100}
= 15{2000 + 2900}
= 15*4900
= Rs. 73500
So, the total amount paid after the 30th installment is Rs. 73500.
The remaining amount of loan = total loan amount - amount paid after the 30th installment
= 118000 - 73500
= Rs. 44500
So, the he still has to pay Rs. 44500 after the payment of the 30th installment.
Answer.
Let us assume that the loan is cleared in 'n' months.
Then, The amounts are in AP with first term (a) = 1000, the common difference (d) = 100 and the Sum of amounts (Sn) = Rs. 118000
So, according to the question.
Sn = n/2{2a + (n - 1)d}
118000 = n/2{2 × 1000 + (n - 1)100}
236000 = n{2000 - 100 + 100n}
236000 = 1900n + 100n²
100n² + 1900n - 236000 = 0
Dividing the above by 100, we get.
n² + 19 - 2360 = 0
n² + 59n - 40n - 2360 = 0
n(n + 59) - 40(n +59) = 0
(n - 40) (n + 59) = 0
n = 40 or n = -59
Since the value of cannot be negative so, n = 40 is correct.
Therefore, the loan will be cleared in 40 months.
Now, the amount to be paid by him in the 30th installment is calculated as under.
a₃₀ = a + 29d where a = 1000 and d = 100
a₃₀ = 1000 + (29*100)
= 1000 + 2900
a₃₀ = 3900
so, the amount to be paid by him in the 30th installment is Rs. 3900
Now, the amount of remaining loan that he has to pay after the 30th installment is calculated as under.
S₃₀ = n/2{2a + (n - 1)d}
= 30/2{2 × 1000 + (30 - 1)100}
= 15{2000 + 2900}
= 15*4900
= Rs. 73500
So, the total amount paid after the 30th installment is Rs. 73500.
The remaining amount of loan = total loan amount - amount paid after the 30th installment
= 118000 - 73500
= Rs. 44500
So, the he still has to pay Rs. 44500 after the payment of the 30th installment.
Answer.
Answered by
84
Answer:
3900
Step-by-step explanation:
Let us assume that the loan is cleared in 'n' months.
Now, the amount to be paid by him in the 30th installment:
a₃₀ = a + 29d where a = 1000 and d = 100
a₃₀ = 1000 + (29*100)
= 1000 + 2900
a₃₀ = 3900
Hence, the amount to be paid by him in the 30th installment is Rs. 3900
S₃₀ = n/2{2a + (n - 1)d}
= 30/2{2 × 1000 + (30 - 1)100}
= 15{2000 + 2900}
= 15*4900
= Rs. 73500
So, the total amount paid after the 30th installment is Rs. 73500.
The remaining amount of loan = total loan amount - amount paid after the 30th installment
= 118000 - 73500
= Rs. 44500
He still has to pay Rs. 44500 after the payment of the 30th installment.
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