Chapter 9: Sequences And Series Miscellaneous Exercise

(i) Given, 5 + 55 + 555 + …

Let S_n = 5 + 55 + 555 + ….. up to n terms

(ii) Given, .6 + .66 + . 666 + …

Let S_n = 06. + 0.66 + 0.666 + … up to n terms

Given series is 2 × 4 + 4 × 6 + 6 × 8 + … n terms

∴ n^th term = a_n = 2n × (2n + 2) = 4n² + 4n

The 20^th term,

a₂₀ = 4 (20)² + 4(20) = 4 (400) + 80 = 1600 + 80 = 1680

Therefore, the 20^th term of the series is 1680.

The given series is 3 + 7 + 13 + 21 + 31 + …

S = 3 + 7 + 13 + 21 + 31 + …+ a_n–1 + a_n

S = 3 + 7 + 13 + 21 + …. + a_{n – 2} + a_{n – 1} + a_n

On subtracting both equations, we get

S – S = [3 + (7 + 13 + 21 + 31 + …+ a_n–1 + a_n)] – [(3 + 7 + 13 + 21 + 31 + …+ a_n–1) + a_n]

S – S = 3 + [(7 – 3) + (13 – 7) + (21 – 13) + … + (a_n – a_n–1)] – a_n

0 = 3 + [4 + 6 + 8 + … (n –1) terms] – a_n

a_n = 3 + [4 + 6 + 8 + … (n –1) terms]

From the question, we have

Therefore, from (1) and (2), we have 9S₂² = S₃ (1 + 8S₁).

n^th term of the numerator = n(n + 1)² = n³ + 2n² + n

n^th term of the denominator = n²(n + 1) = n³ + n²

Given, the farmer pays Rs 6000 in cash.

So, the unpaid amount = Rs 12000 – Rs 6000 = Rs 6000

From the question, the interest paid annually will be,

12% of 6000, 12% of 5500, 12% of 5000, …, 12% of 500

Hence, the total interest to be paid = 12% of 6000 + 12% of 5500 + 12% of 5000 + … + 12% of 500

= 12% of (6000 + 5500 + 5000 + … + 500)

= 12% of (500 + 1000 + 1500 + … + 6000)

It’s seen that the series 500, 1000, 1500 … 6000 is an A.P. with the first term and common difference both equal to 500.

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

So, 6000 = 500 + (n – 1) 500

1 + (n – 1) = 12

n = 12

Now,

The sum of the A.P = 12/2 [2(500) + (12 – 1)(500)] = 6 [1000 + 5500] = 6(6500) = 39000

Hence, the total interest to be paid = 12% of (500 + 1000 + 1500 + … + 6000)

= 12% of 39000 = Rs 4680

Therefore, the tractor will cost the farmer = (Rs 12000 + Rs 4680) = Rs 16680

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

From 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 the 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:

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

It’s seen that,

The numbers of letters mailed forms a G.P.: 4, 4², … 4⁸

Here, first term = 4 and common ratio = 4

And the number of terms = 8

The sum of n terms of a G.P. is given by:

Also, given that the cost to mail one letter is 50 paisa.

Hence, Cost of mailing 87380 letters = Rs 87380 x (50/100) = Rs 43690 = Rs 43690

Therefore, the amount spent when 8^th set of letter is mailed will be Rs 43690.

Given, the man deposited Rs 10000 in a bank at the rate of 5% simple interest annually.

Hence, the interest in first year = (5/100) x Rs 10000 = Rs 500

So, the amount in 15^th year = Rs

= Rs 10000 + 14 × Rs 500

= Rs 10000 + Rs 7000

= Rs 17000

And, the amount after 20 years =

= Rs 10000 + 20 × Rs 500

= Rs 10000 + Rs 10000

= Rs 20000

Therefore, the amount in the 15^th year is Rs 17000, and the total amount after 20 years will be Rs 20000.