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 \(\( a \)\).
- The second installment will be \(\( 3a \)\).
- The third installment will be \(\( 3^2a = 9a \)\).
- The fourth installment will be \(\( 3^3a = 27a \)\).
- The fifth installment will be \(\( 3^4a = 81a \)\).
- The sixth installment will be \(\( 3^5a = 243a \)\).
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:
\(\[ S = a \frac{r^n - 1}{r - 1} \]\)
where:
- \(\( S \)\) is the sum (Rs. 43,680),
- \(\( a \)\) is the first term,
- \(\( r \)\) is the common ratio (3 in this case),
- \(\( n \)\) is the number of terms (6).
Plugging in the values, we have:
\(\[ 43680 = a \frac{3^6 - 1}{3 - 1} \]\)
\(\[ 43680 = a \frac{729 - 1}{3 - 1} \]\)
\(\[ 43680 = a \frac{728}{2} \]\)
\(\[ 43680 = 364a \]\)
Solving for \(\( a \)\):
\(\[ a = \frac{43680}{364} \]\)
\(\[ a = 120 \]\)
So, the first installment \(\( a \)\) is Rs. 120.
The last installment is the sixth installment, which is \(\( 3^5a \)\):
\(\[ 3^5 = 243 \]\)
\(\[ 243 \times 120 = 29160 \]\)
So, the last installment is Rs. 29,160.
Finally, the difference between the first and the last installments is:
\(\[ 29160 - 120 = 29040 \]\)
Thus, the difference between the first and the last installments is Rs. 29,040.
