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Question
If ₹ 3900 will have to be repaid in 12 monthly instalments such that each instalment being more than the preceding one by ₹ 10, then find the amount of the first and last instalment.
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Solution
The instalments are in A.P.
Amount repaid in 12 instalments (S12) = 3900
Number of instalments (n) = 12
Each instalment is more than the preceding one by ₹ 10.
∴ d = 10
Now, `S_n = n/2 [2a + (n - 1)d]`
∴ `S_12 = 12/2[2a + (12 - 1)(10)]`
∴ 3900 = 6[2a + 11(10)]
∴ 3900 = 6(2a + 110)
∴ `3900/6` = 2a + 110
∴ 650 = 2a + 110
∴ 2a = 540
∴ a = `540/2`
∴ a = 270
tn = a + (n – 1)d
∴ t12 = 270 + (12 – 1)(10)
= 270 + 11(10)
= 270 + 110
= 380
∴ Amount of the first instalment is ₹ 270 and that of the last instalment is ₹ 380.
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