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, T₆ = 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 S₁₅  = [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: T₁₀  =  3425 + (10 – 1) (- 200)

= 3425 – 9 x 200 = 3425 -1800 = 1625

And the value of 11^th EMI: T₁₁  =  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.