Question

Rs3900 is repaid in 12 EMI. Every EMI amount is 10 Rs increased by its previous EMI then
Find 1st and last EMI amount?

Answer 1

Given info : Rs3900 is repaid in 12 EMI. Every EMI amount is 10 Rs increased by its previous EMI.

To find : The 1st and last EMI amount are ...

solution : let first EMI is x

a/c to question, Every EMI amount is 10 Rs increased by its previous EMI.

it means, EMI are in arithmetic progression where common difference is 10 Rs.

so sum of n ap terms , Sn = n/2 [2a + (n - 1)d]

here sum of 12 EMIs , S₁₂ = 3900 Rs.

n = 12, d = 10

⇒3900 = 12/2[2a + (12 - 1)10]

⇒650 = 2a + 110

⇒540 = 2a

⇒a = 270

Therefore the first EMI is 270 Rs.

now last EMI = 12th term = a + (12 - 1)d

= a + 11d = 270 + 11 × 10 = 380 Rs.

Therefore the last EMI is 380 Rs.

Answer 2

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• let first EMI is x

according to the quéstion

Every EMI amount is 10 Rs increased by its previous EMI.

it means, EMI are in arithmetic progression where common difference is 10 Rs.

so sum of n ap terms , Sn = n/2 [2a + (n - 1)d]

here sum of 12 EMIs , S₁₂ = 3900 Rs.

n = 12, d = 10

⇒3900 = 12/2[2a + (12 - 1)10]

⇒650 = 2a + 110

⇒540 = 2a

⇒a = 270

Therefore the first EMI is 270 Rs.

now last EMI = 12th term = a + (12 - 1)d

= a + 11d = 270 + 11 × 10 = 380 Rs.

Therefore the last EMI is 380 Rs.