(i) 2, 7, 12, . . ., to 10 terms.

(ii) –37, –33, –29, . . ., to 12 terms.

(iii) 0.6, 1.7, 2.8, . . ., to 100 terms.

(iv) 1/15, 1/12, 1/10, . . ., to 11 terms.

**Solution:**

**(i) Given, 2, 7, 12,…, to 10 terms**

First term, a = 2

And common difference, d = a_{2} − a_{1} = 7 − 2 = 5

n = 10

We know that the formula for the sum of the nth term in the AP series is,

S_{n} = n/2 [2a + (n – 1)d]

S_{10} = 10/2 [2(2) + (10 – 1) × 5]

= 5[4 + (9) × (5)]

= 5 × 49 = 245

**(ii) Given, −37, −33, −29,…, to 12 terms**

First term, a = −37

And common difference, d = a_{2}− a_{1}

d = (−33)−(−37)

= − 33 + 37 = 4

n = 12

We know that the formula for the sum of the nth term in the AP series is,

S_{n} = n/2 [2a + (n – 1)d]

S_{12} = 12/2 [2(-37) + (12 – 1) × 4]

= 6[-74 + 11 × 4]

= 6[-74 + 44]

= 6(-30) = -180

**(iii) Given, 0.6, 1.7, 2.8,…, to 100 terms**

First term, a = 0.6

Common difference, d = a_{2} − a_{1} = 1.7 − 0.6 = 1.1

n = 100

We know that the formula for the sum of the nth term in the AP series is,

S_{n} = n/2[2a + (n – 1)d]

S_{12} = 50/2 [1.2 + (99) × 1.1]

= 50[1.2 + 108.9]

= 50[110.1]

= 5505

**(iv) Given, 1/15, 1/12, 1/10, …… , to 11 terms**

First term, a = 1/5

Common difference, d = a_{2 }– a_{1} = (1/12) – (1/5) = 1/60

And the number of terms n = 11

We know that the formula for the sum of the nth term in the AP series is,

S_{n} = n /2 [2a + ( n – 1) d ]

S_{n} = (11/2) [2(1/15) + (11 – 1) × 1/ 60]

= 11/2(2/15 + 10/60)

= 11/2 (9/30)

= 33/20