Jaspal Singh repays his total loan of Rs. 118 | vedclass

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Question

(A) The instalments form an Arithmetic Progression $(AP)$ where the first term $a = 1000$ and the common difference $d = 100$.$1$. The $30^{	ext{th}}

Vedclass · Accepted Answer

$Rs. 3900$

Vedclass · Answer

(A) The instalments form an Arithmetic Progression $(AP)$ where the first term $a = 1000$ and the common difference $d = 100$.$1$. The $30^{	ext{th}}$ instalment is given by the formula $a_n = a + (n - 1)d$.For $n = 30$,$a_{30} = 1000 + (30 - 1) 	imes 100 = 1000 + 2900 = Rs. 3900$.$2$. The total amount paid in $30$ instalments is given by the sum formula $S_n = \frac{n}{2} [2a + (n - 1)d]$.$S_{30} = \frac{30}{2} [2(1000) + (30 - 1) 	imes 100] = 15 [2000 + 2900] = 15 	imes 4900 = Rs. 73500$.$3$. The remaining loan amount after $30$ instalments is $Total Loan - S_{30} = 118000 - 73500 = Rs. 44500$.

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