Sum

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 (S_{12}) = 3900

Number of instalments (n) = 12

Each instalment is more than the preceding one by ₹ 10.

∴ d = 10

Now, S_{n} = `"n"/2 [2"a" + ("n" - 1)"d"]`

∴ S_{12} = `12/2[2"a" + (12 - 1)(10)]`

∴ 3900 = 6[2a + 11(10)]

∴ 3900 = 6(2a + 110)

∴ `3900/6` = 2a + 110

∴ 650 = 2a + 110

∴ 2a = 540

∴ a = `540/2` = 270

t_{n} = a + (n – 1)d

∴ t_{12} = 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.

Concept: Sum of First n Terms of an AP

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