EX. NO. 4

1. Find the sum of the first n natural numbers and hence find the sum of first 20 natural numbers.

Sol. The first n natural numbers are 1, 2, 3, 4, ......nThese natural number form an A.P. with a = 1, d = 1t1 = 1 and tn = nAlternative method :Sn = n/2 [t1 + tn]S20 = 20/2[t1 + t20]= 10 [1 + 20]= 10 [21]S20 = 210The sum of first twenty terms is 210.

2. Find the sum of all odd natural numbers from 1 to 150. Sol. The odd natural numbers from 1 to 150 are 1, 3, 5, 7, 9, .........., 149.These numbers form an A.P. with a = 1, d = 2Let, 149 be nth term of an A.P.tn = a + (n – 1) d149 = 1 + (n – 1) 2 = 1 + 2n – 2 = 2n – 1149 + 1 = 2n2n = 150n = 75 ∴ 149 is 75th term of A.P.

Sn = n/2 [t1 + tn]

S75 = 75/2[t1 + t75]

= 75/2[ [1 + 149]

= 75/2[ [150]

S75 = 75 x 75 =5625

3. Find S10 if a = 6 and d = 3.

Sol. For an A.P. a = 6, d = 3Sn = n/2 [2a + (n – 1)d]S10 = 10/2 [2(6) + (10 – 1 ) 3]S10 = 5 [2 (6) + 9 (3)]S10 = 5 (12 + 27)S10 = 5 (39)S10 = 195

4. Find the sum of all numbers from 1 to 140 which are divisible by 4.Sol. The natural numbers from 1 to 140 that are divisible by 4 are 4, 8, 12, 16, .............., 140Here, a = 4, d = t2 – t1 = 8 – 4 = 4 and tn = 140tn = a + (n – 1) d∴ 140 = 4 + (n – 1) 4∴ 140 = 4 + 4n – 4∴ 140 = 4n∴ n = 140 /4

∴ n = 35

Sn = n/2 [2a + (n – 1)d]∴ S35 = 35/2 [2a + (35 – 1 ) d]∴ S35 = 35/2 [2 (4) + 34 (4)]∴ S35 = 35/2 (8 + 136)∴ S35 = 35/2 (144)∴ S35 = 35 x 72∴ S35 = 2520

5. Find the sum of the first n odd natural numbers. Hence find 1 + 3 + 5 + ... + 101. Sol. The first n odd natural numbers are 1, 3, 5, 7, ............., nHere, a = 1, d = t2 – t1 = 3 – 1 = 2Sn = n/2 [2a + (n – 1)d]∴ Sn = n/2 [2(1) + (n – 1)(2)]∴ Sn = n/2 (2 + 2n – 2)∴ Sn = n/2 (2n)∴ Sn = n2 ........ eq. (1)

Now, we have 1 + 3 + 5 + ........ + 101Let, tn = 101tn = a + (n – 1) d∴ 101 = a + (n – 1) d∴ 101 = 1 + (n – 1) 2∴ 101 = 1 + 2n – 2∴ 101 = 2n – 1∴ 101 + 1 = 2n∴ 2n = 102∴ n = 102/2

∴ n = 51

Now, 101 is the 51st term of A.P.,We have to find sum of 51 terms i.e. S51,Sn = n2 [From (i)]∴ S51 = (51)2

∴ S51 = 2601

6. Obtain the sum of the 56 terms of an A. P. whose 19th and 38th terms are52 and 148 respectively.

Sol. t19 = 52, t38 = 148tn = a + (n – 1) d∴ t19 = a + (19 – 1) d∴ 52 = a + 18d ......(i)∴ t38 = a + (38 – 1) d∴ 148 = a + 37d ......(ii)

Adding (i) and (ii) we get,a + 18d = 52a + 37d = 148 .2a + 55d = 200 . ..........Eq. no. (iii)Sn = n/2[2a + (n – 1)d]∴ S56 = 56/2 [2a + (56 – 1) d]∴ S56 = 28[2a + 55d]∴ S56 = 28[200] [from eq. no. (iii)∴ S56 = 5600∴ Sum of first 56 terms of A.P. is 5600.

7. The sum of the first 55 terms of an A. P. is 3300. Find the 28th term.Sol. S55 = 3300 [Given]Sn = n/2[2a + (n – 1)d]∴ S55 = 55/2[2a + (55 – 1) d]∴ 3300 = 55/2[2a + 54d]∴ 3300 = 55/2 × (2)[a + 27d]∴ 3300 = 55[a + 27d]3300/55= a + 27d∴ a + 27d= 60 ......eq. no. (1)

But, tn = a + (n – 1) d∴ t28 = a + (28 – 1) d∴ t28 = a + 27d∴ t28 = 60 [From (i)]∴ Twenty eighth term of A.P. is 60.

8. Find the sum of the first n even natural numbers. Hence find the sum of first 20 even natural numbers. Sol. The first n even natural numbers are 2, 4, 6, 8, .....Here, a = 2, d = t2 – t1 = 4 – 2 = 2∴ Sn = n/2[2a + (n – 1)d]∴ Sn = n/2[2 (2) + (n – 1) 2]∴ Sn = n/2[4 + 2n – 2]∴ Sn = n/2[2n + 2]∴ Sn =n/2 ×(2) (n + 1)∴ Sn = n (n + 1)

∴ S20 = 20 (20 + 1)∴ S20 = 20 (21)∴ S20 = 420∴ Sum of first twenty even natural numbers is 420