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Question
A man arranges to pay off a debt of Rs 3600 by 40 annual instalments which form an arithmetic series. When 30 of the instalments are paid, he dies leaving one-third of the debt unpaid, find the value of the first instalment.
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Solution
Let
\[S_{40}\] denote the total loan amount to be paid in 40 annual instalments.
∴ \[S_{40}\] = 3600
Let Rs a be the value of the first instalment and Rs d be the common difference.
We know:
\[S_n = \frac{n}{2}\left\{ 2a + \left( n - 1 \right)d \right\}\]
\[\Rightarrow \frac{40}{2}\left\{ 2a + \left( 40 - 1 \right)d \right\} = 3600\]
\[ \Rightarrow 20\left\{ 2a + 39d \right\} = 3600\]
\[ \Rightarrow 2a + 39d = 180 . . . . . . \left( 1 \right)\]
\[\text { Also, } S_{40} - S_{30} = \frac{1}{3} \times 3600\]
\[ \Rightarrow 3600 - S_{30} = 1200\]
\[ \Rightarrow S_{30} = 2400\]
\[ \Rightarrow \frac{30}{2}\left\{ 2a + \left( 30 - 1 \right)d \right\} = 2400\]
\[ \Rightarrow 15\left\{ 2a + 29d \right\} = 2400\]
\[ \Rightarrow 2a + 29d = 160 . . . . . \left( 2 \right)\]
On solving equations \[\left( 1 \right) \text { and } \left( 2 \right)\]
we get:
d = 2 and a =51
Hence, the value of the first instalment is Rs 51
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