(i) 7 + 10 ½ + 14 + . . . + 84

(ii) 34 + 32 + 30 + . . . + 10

(iii) –5 + (–8) + (–11) + . . . + (–230)

**Solution:**

**(i) 7 + 10 ½ + 14 + . . . + 84**

First term, a = 7

n^{th} term, a_{n }= 84

Common difference, d = a_{2} – a_{1} = 10 ½ – 7 = (21/2) – 7 = 7/2

Let 84 be the n^{th }term of this AP, then as per the n^{th} term formula,

a_{n }= a + (n – 1)d

84 = 7 + (n – 1) × 7/2

77 = (n – 1) × 7/2

22 = n − 1

n = 23

We know that, sum of n term is;

S_{n} = n/2 (a + l) , where l = 84

S_{n} = 23/2 (7 + 84)

S_{n} = (23 × 91/2) = 2093/2

S_{n} = 1046 ½

**(ii) 34 + 32 + 30 + ……….. + 10**

For this A.P.,

first term, a = 34

common difference, d = a_{2}− a_{1} = 32 − 34 = −2

n^{th} term, a_{n}= 10

Let 10 be the n^{th} term of this AP, therefore,

a_{n }= a + (n − 1)d

10 = 34 + (n − 1)(−2)

−24 = (n − 1)(−2)

12 = n −1

n = 13

We know that the sum of n terms is;

S_{n} = n/2 (a + l), where l = 10

= 13/2 (34 + 10)

= (13 × 44/2) = 13 × 22

= 286

**(iii) (−5) + (−8) + (−11) + ………… + (−230)**

For this A.P.,

First term, a = −5

nth term, a_{n}= −230

Common difference, d = a_{2 }− a_{1} = (−8) − (−5)

⇒d = −8 + 5 = −3

Let −230 be the n^{th} term of this AP, and by the n^{th} term formula we know,

a_{n} = a + (n − 1) d

−230 = −5 + (n − 1)(−3)

−225 = (n − 1)(−3)

(n − 1) = 75

n = 76

And, Sum of n term,

S_{n} = n /2 ( a + l )

= 76/2 [(-5) + (-230)]

= 38(-235)

= -8930