**Q) **Deepa has to buy a scooty. She can buy scooty either making cashdown payment of ₹ 25,000 or by making 15 monthly instalments as below.

Ist month – ₹ 3425, IInd month – ₹ 3225, IIIrd month – ₹ 3025, IVth month – ₹ 2825 and so on

i. Find the amount of 6th instalment.

ii. Total amount paid in 15 instalments.

OR

ii. If Deepa pays ₹2625 then find the number of instalment.

iii. Deepa paid 10th and 11th instalment together find the amount paid that month.

**Ans: **Here, we are given following data:

As we can clearly see that the value of monthly payment (EMI) is going down by Rs. 200 every month.

Hence, this will make an Arithematic Progression (AP), with a = 3425 and d = – 200

**(i) Amount of 6 ^{th} Installment:**

We know that the n^{th} term of an AP is given by: T_{n } = a + (n-1) d

Therefore, value of 6^{th} term, T_{6 }= 3425 + (6 – 1) x (- 200) = 2425

**Therefore, Deepa’s 6 ^{th} installment will be Rs. 2,425/-**

*(Quick Check: 4th installment is 2825, Each EMI is going down by 200/- hence, 6th EMI value will be 2825 – 2 x 200 = 2825 – 400 = 2425)*

**(ii) A. Total Amount paid in 15 installments:**

We know that the sum of n terms of an AP is given by: S_{n }_{ }= [2a + (n-1) d]

Hence, sum of 15 terms S_{15 }_{ }= [2 x 3425 + (15 – 1) (- 200)] = 15 x 4825 = 30,375

**Therefore, Total Rs. 30,375 will be paid in 15 installments.**

**OR**

**(ii) B. Installment number for amount of Rs. 2,625:**

We know that the n^{th} term of an AP is given by: T_{n } = a + (n-1) d

Here, value of nth term is given to us, and we need to find value of n.

Hence, 2625 = 3425 + (n – 1) x (- 200)

– 800 = -200 (n – 1)

or n – 1 = 4

n = 5

**Therefore, 5 ^{th} installment amount will be Rs. 2,625/-**

**iii. Total amount of 10 ^{th} and 11^{th} instalments**

We know that the n^{th} term of an AP is given by: T_{n } = a + (n – 1) d

Hence, the value of 10^{th} EMI: T_{10 } = 3425 + (10 – 1) (- 200)

= 3425 – 9 x 200 = 3425 -1800 = 1625

And the value of 11^{th} EMI: T_{11 } = 3425 + (11 – 1) (- 200)

= 3425 – 10 x 200 = 3425 – 2000 = 1425

Hence, Amount paid will be sum of 10^{th} EMI + 11^{th} EMI = 1625 + 1425 = 3050

**Therefore, Deepa will pay total Rs. 3,050 as sum of 10 ^{th} and 11^{th} instalment together.**