### __10th Maths- Chapter 2 – Numbers and Sequences Exercise 2.6 (All Sums)__

**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 = t2 – t1 = 7 – 3 = 4

n = 40

Sn = n/2 (2a + (n – 1)d)

S40 = 20/2 (2× 3 + 39d)

= 20(6 + 39 × 4)

= 20(6 + 156)

= 20 × 162

= 3240

(ii) 102, 97, 952,… up to 27 terms

a = 102,

d = t2 – t1

= 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?**

**Solution:**

5,7,9, 11, 13,…

Sn = 480

a = 5, d = 2, Sn = 480

2n2 + 8n – 960 = 0

⇒ n2 + 4n – 480 = 0

⇒ n2 + 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 nth term is 4n – 3.**

**Solution:**

Number of terns (n) = 28

tn = 4n – 3

t1 = 4(1) – 3 = 4 – 3 = 1

t2 = 4(2) – 3 = 8 – 3 = 5

t3 = 4(3) – 3 = 12 – 3 = 9

Here a = 1, d = 5 – 1 = 4

S28 = n/2 [2a + (n – 1)d]

= 28/2 [2 + (27) (4)]

= 14 [2 + 108]

= 14 × 110

= 1540

Sum of 28 terms = 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 Sn = 2n2 – 3n

S1 = 2(1)2 – 3(1) = 2 – 3 = – 1

⇒ t1 = a = – 1

S2 = 2(22) – 3(2) = 8 – 6 = 2

t2 = S2 – S1 = 2 – (-1) = 3

∴ d = t2 – t1 = 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 104th term and 4th term of an A.P are 125 and 0. Find the sum of first 35 terms.**

**Solution:**

t104 = 125

t4 = 0

a + (n – 1)d = tn

**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 = n2.

We can use the formula n2 = 2252

= 50625

**Question 7.**

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

**Solution:**

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 making 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 pay in installments.**

**Solution:**

(i) Amount paid in 10 installments

4800 + 4750 + 4700 + ……………. 10

Here a = 4800; d = – 50 ; n = 10

Sn = n/2 [2a +(n – 1)d]

S10 = 10/2 [2 × 4800 +9(-50)]

= 10/2 [9600 – 450]

= 5 [9150]

= 45750

Amount paid in 10 installments

= ₹45750

(ii) Extra amount paid = amount paid in 10

installment – cost of the laptop

= ₹45750 – 40,000

= ₹ 5750

(i) Amount paid in 10 installments = ₹ 45750

(ii) Difference in payment = ₹ 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

Sn = 65000

Sn = n/2 (2a + (n – 1)d)

= 65000

n/2(2 × 400 + (n – 1)300) = 65000

n(800 + 300n – 300) = 130000

n(500 + 300n) = 130000

500n + 300n2 = 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:**

Total number of steps = 30

∴ n = 30

Number of bricks for the bottom = 100

a = 100

2 bricks is less for each step

(i) Number of bricks required for the top most step

tn = a + (n – 1)d

t30 = 100 + 29 (-2)

= 100 – 58

= 42

(ii) Number of bricks required

Sn = n/2 [2a + (n-1) d]

S30 = 30/2 [200 + 29 (-2)]

= 15[200 – 58]

= 2130

(i) Number of bricks required for the top most step = 42 bricks

(ii) Number of bricks required = 2130

**Question 11.**

**If S1,S2,S3 , …. , Sm 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**

**S1 + S2 + S3 + …. + Sm = 1/2 mn(mn + 1).**

**Solution:**

**Question 12.**

**Find the sum**

Solution: