## Tamilnadu State Board Class 10 Maths Solutions Chapter 2 Numbers and Sequences Ex 2.6

Question 1.

Find the sum of the following

(i) 3,7,11,… up to 40 terms.

(ii) 102, 97, 92,… up to 27 terms.

(iii) 6 + 13 + 20 + … + 97

Solution:

(i) 3, 7, 11,. . . upto 40 terms.

a = 3, d = t_{2} – t_{1} = 7 – 3 = 4

n = 40

(ii) 102, 97, 952,… up to 27 terms

a = 102,

d = t_{2} – t_{1}

= 97 – 102 = -5

n = 27

(iii) 6 + 13 + 20 + … + 97

a = 6,d = 7,l = 97

Question 2.

How many consecutive odd integers beginning with 5 will sum to 480?

Answer:

5,7,9, 11, 13,…

s_{n} = 480

a = 5, d = 2, s_{n} = 480

2 n^{2} + 8 n – 960 = 0

⇒ n^{2} + An – 480 = 0

⇒ n^{2} + 24n – 20n – 480 = 0

⇒ n(n + 24) – 20(n + 24) = 0

⇒ (n – 20)(n + 24) = 0

⇒ n = 20,-24

No. of terms cannot be -ve.

∴ No. of consecutive odd integers beginning with 5 will sum to 480 is 20.

Question 3.

Find the sum of first 28 terms of an A.P. whose n^{th} term is 4n – 3.

Solution:

n = 28

t_{n} = 4n – 3

t_{1} = 4 × 1 – 3 = 1

t_{2} = 4 × 2 – 3 = 5

t_{28} = 4 × 28 – 3

= 112 – 3 = 109

∴ a = 1, d = t_{2} – t_{1} = 5 – 1 = 4

l = 109.

S_{n} = \(\frac { n }{ 2 } \) (2a+(n-1)d)

S_{28} = \(\frac { 28 }{ 2 } \) (2 × 1 + 27 × 4)

= 14(2 + 108)

= 14 × 110

= 1540

Question 4.

The sum of first n terms of a certain series is given as 2n2 – 3n. Show that the series is an A.P.

Solution:

Given S_{n} = 2n^{2} – 3n

S_{1} = 2(1)^{2} – 3(1) = 2 – 3 = – 1

⇒ t_{1} = a = – 1

S_{2} = 2(2^{2}) – 3(2) = 8 – 6 = 2

t_{2} = S_{2} – S_{1} = 2 – (-1) = 3

∴ d = t_{2} – t_{1} = 3 – (-1) = 4

Consider a, a + d, a + 2d,….

-1, -1 + 4, -1 + 2(4),…

-1, 3, 7,….

Clearly this is an A.P with a = – 1, and d = 4.

Question 5.

The 104^{th} term and 4th term of an A.P are 125 and 0. Find the sum of first 35 terms.

Solution:

t_{104} = 125

t_{4} = 0

a + (n – 1)d = t_{n
}

Question 6.

Find the sum of all odd positive integers less than 450.

Solution:

Sum of all odd positive integers less than 450 is given by

1 + 3 + 5 + … + 449

a = 1

d = 2

l = 449

= 50625

Another method:

Sum of all +ve odd integers = n^{2}.

We can use the formula n^{2} = 225^{2}

= 50625

Question 7.

Find the sum of all natural numbers between 602 and 902 which are not divisible by 4.

Answer:

Natural numbers between 602 and 902

603,604, …, 901

a = 603, l = 901, d = 1,

Sum of all natural numbers between 602 and 902 which are not divisible by 4.

= Sum of all natural numbers between 602 and 902

= Sum of all natural numbers between 602 and 902 which are divisible by 4.

l = 902 – 2 = 900

To make 602 divisible by 4 we have to add 2 to 602.

∴ 602 + 2 = 604 which is divisible by 4.

To make 902 divisible by 4, subtract 2 from 902.

∴ 900 is the last number divisible by 4.

Sum of all natural numbers between 602 and 902 which are not divisible 4.

= 224848 – 56400

= 168448

Question 8.

Raghu wish to buy a laptop. He can buy it by paying ₹40,000 cash or by giving it in 10 installments as ₹4800 in the first month, ₹4750 in the second month, ₹4700 in the third month and so on. If he pays the money in this fashion, find

(i) total amount paid in 10 installments.

(ii) how much extra amount that he has to pay than the cost?

Solution:

4800 + 4750 + 4700 + … 10 terms

Here a = 4800

(i) d = t_{2} – t_{1} = 4750 – 4800 = -50

n = 10

S_{n} = \(\frac { n }{ 2 } \) (2a + (n – 1)d)

s_{10} = \(\frac { 10 }{ 2 } \) (2 × 4800 + 9 × -50)

= 5 (9600 – 450)

= 5 × 9150 = 45750

Total amount paid in 10 installments = ₹ 45750.

(ii) The extra amount he pays in installments

= ₹ 45750 – ₹ 40,000

= ₹ 5750

Question 9.

A man repays a loan of ₹65,000 by paying ₹400 in the first month and then increasing the payment by ₹300 every month. How long will it take for him to clear the loan?

Solution:

Loan Amount = ₹ 65,000

Repayment through installments

400 + 700 + 1000 + 1300 + …

a = 400

d = 300

S_{n} = 65000

S_{n} = \(\frac { n }{ 2 } \) (2a+(n – 1)d)

= 65000

\(\frac { n }{ 2 } \) (2 × 400 + (n – 1)300) = 65000

n(800 + 300n – 300) = 130000

n(500 + 300n) = 130000

500n + 300n^{2} = 130000

Number of terms should be (+ve) and cannot be (-ve) or fractional number.

∴ He will take 20 months to clear the loans.

Question 10.

A brick staircase has a total of 30 steps. The bottom step requires 100 bricks. Each successive step requires two bricks less than the previous step.

(i) How many bricks are required for the top most step?

(ii) How many bricks are required to build the stair case?

Solution:

100 + 98 + 96 + 94 + … 30 steps.

Here

a = 100

d = -2

n = 30

∴ S_{n} = \(\frac { n }{ 2 } \) (2a + (n – 1)d)

S_{30} = \(\frac { 30 }{ 2 } \) (2 × 100 + 29 × -2)

= 15(200 – 58)

= 15 × 142

= 2130

t_{30} = a + (n – 1)d

= 100 + 29 × -2

= 100 – 58

= 42

(i) No. of bricks required for the top step are 42.

(ii) No. of bricks required to build the stair case are 2130.

Question 11.

If S_{1},S_{2},S_{3} , …. , S_{m} are the sums of n terms of m A.P.’s whose first terms are 1,2,3,…,m and whose common differences are 1, 3, 5,…, (2m – 1) respectively, then show that

S_{1} + S_{2} + S_{3} + …. + Sm = \(\frac { 1 }{ 2 } \) mn(mn + 1).

Solution:

Question 12.

Find the sum

Solution: