Answer :
Sure, let's solve the problem step by step.
Sarita borrowed Rs. 43,680 from her friend Garima with the condition that she would pay it back in 6 installments, where each installment is three times the previous one. This indicates that the installments form a geometric series.
First, let's define the installments:
- Let the first installment be [tex]\( a \)[/tex].
- The second installment will be [tex]\( 3a \)[/tex].
- The third installment will be [tex]\( 3^2a = 9a \)[/tex].
- The fourth installment will be [tex]\( 3^3a = 27a \)[/tex].
- The fifth installment will be [tex]\( 3^4a = 81a \)[/tex].
- The sixth installment will be [tex]\( 3^5a = 243a \)[/tex].
The sum of these installments equals the total borrowed amount, which is Rs. 43,680.
The sum of a geometric series can be calculated using the formula:
[tex]\[ S = a \frac{r^n - 1}{r - 1} \][/tex]
where:
- [tex]\( S \)[/tex] is the sum (Rs. 43,680),
- [tex]\( a \)[/tex] is the first term,
- [tex]\( r \)[/tex] is the common ratio (3 in this case),
- [tex]\( n \)[/tex] is the number of terms (6).
Plugging in the values, we have:
[tex]\[ 43680 = a \frac{3^6 - 1}{3 - 1} \][/tex]
[tex]\[ 43680 = a \frac{729 - 1}{3 - 1} \][/tex]
[tex]\[ 43680 = a \frac{728}{2} \][/tex]
[tex]\[ 43680 = 364a \][/tex]
Solving for [tex]\( a \)[/tex]:
[tex]\[ a = \frac{43680}{364} \][/tex]
[tex]\[ a = 120 \][/tex]
So, the first installment [tex]\( a \)[/tex] is Rs. 120.
The last installment is the sixth installment, which is [tex]\( 3^5a \)[/tex]:
[tex]\[ 3^5 = 243 \][/tex]
[tex]\[ 243 \times 120 = 29160 \][/tex]
So, the last installment is Rs. 29,160.
Finally, the difference between the first and the last installments is:
[tex]\[ 29160 - 120 = 29040 \][/tex]
Thus, the difference between the first and the last installments is Rs. 29,040.
Sarita borrowed Rs. 43,680 from her friend Garima with the condition that she would pay it back in 6 installments, where each installment is three times the previous one. This indicates that the installments form a geometric series.
First, let's define the installments:
- Let the first installment be [tex]\( a \)[/tex].
- The second installment will be [tex]\( 3a \)[/tex].
- The third installment will be [tex]\( 3^2a = 9a \)[/tex].
- The fourth installment will be [tex]\( 3^3a = 27a \)[/tex].
- The fifth installment will be [tex]\( 3^4a = 81a \)[/tex].
- The sixth installment will be [tex]\( 3^5a = 243a \)[/tex].
The sum of these installments equals the total borrowed amount, which is Rs. 43,680.
The sum of a geometric series can be calculated using the formula:
[tex]\[ S = a \frac{r^n - 1}{r - 1} \][/tex]
where:
- [tex]\( S \)[/tex] is the sum (Rs. 43,680),
- [tex]\( a \)[/tex] is the first term,
- [tex]\( r \)[/tex] is the common ratio (3 in this case),
- [tex]\( n \)[/tex] is the number of terms (6).
Plugging in the values, we have:
[tex]\[ 43680 = a \frac{3^6 - 1}{3 - 1} \][/tex]
[tex]\[ 43680 = a \frac{729 - 1}{3 - 1} \][/tex]
[tex]\[ 43680 = a \frac{728}{2} \][/tex]
[tex]\[ 43680 = 364a \][/tex]
Solving for [tex]\( a \)[/tex]:
[tex]\[ a = \frac{43680}{364} \][/tex]
[tex]\[ a = 120 \][/tex]
So, the first installment [tex]\( a \)[/tex] is Rs. 120.
The last installment is the sixth installment, which is [tex]\( 3^5a \)[/tex]:
[tex]\[ 3^5 = 243 \][/tex]
[tex]\[ 243 \times 120 = 29160 \][/tex]
So, the last installment is Rs. 29,160.
Finally, the difference between the first and the last installments is:
[tex]\[ 29160 - 120 = 29040 \][/tex]
Thus, the difference between the first and the last installments is Rs. 29,040.
