Given, Shamshad Ali buys a scooter for Rs 22000 and pays Rs 4000 in cash.

So, the unpaid amount = Rs 22000 – Rs 4000 = Rs 18000

Form the question, it’s understood that the interest paid annually is

10% of 18000, 10% of 17000, 10% of 16000 … 10% of 1000

Hence, the total interest to be paid = 10% of 18000 + 10% of 17000 + 10% of 16000 + … + 10% of 1000

= 10% of (18000 + 17000 + 16000 + … + 1000)

= 10% of (1000 + 2000 + 3000 + … + 18000)

It’s seen that, 1000, 2000, 3000 … 18000 forms an A.P. with first term and common difference both equal to 1000.

Let’s take the number of terms to be n.

So, 18000 = 1000 + (n – 1) (1000)

n = 18

Now, the sum of the A.P is given by:

\( 1000+2000+.....+18000\)

= \( s_n=\dfrac{18}{2}[2(1000)+(18-1)(1000)]\)

= \( 9[2000+17000]\)

= 171000

Thus,

Total interest paid = 10% of (18000 + 17000 + 16000 + … + 1000)

= 10% of Rs 171000 = Rs 17100

Therefore, the cost of scooter = Rs 22000 + Rs 17100 = Rs 39100

Answered by Abhisek | 1 year agoConstruct a quadratic in x such that A.M. of its roots is A and G.M. is G.

Find the geometric means of the following pairs of numbers:

**(i) **2 and 8

**(ii) **a^{3}b and ab^{3}

**(iii) **–8 and –2

Insert 5 geometric means between \( \dfrac{32}{9}\) and \( \dfrac{81}{2}\).