Question -
Answer -
(ii) 34 + 32 + 30 +……….. + 10
(iii) − 5 + (− 8) + (− 11) + ………… + (− 230)
Solution 
First term, a = 7
nth term, an =84
Let 84 be the nth termof this A.P., then as per the nth term formula,
an = 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;
Sn = n/2(a + l) , l = 84
Sn = 23/2 (7+84)
Sn = (23×91/2) = 2093/2
Solution
(ii) Given, 34 + 32 + 30 + ……….. + 10
For this A.P.,
first term, a = 34
common difference, d = a2−a1 =32−34 = −2
nth term, an= 10
Let 10 be the nth termof this A.P., therefore,
an= a +(n−1)d
10 = 34+(n−1)(−2)
−24 = (n −1)(−2)
12 = n −1
n = 13
We know that, sum of n terms is;
Sn = n/2 (a +l) , l= 10
= 13/2 (34 + 10)
= (13×44/2) = 13 × 22
= 286
Solution
(iii) Given, (−5) + (−8) + (−11) + ………… + (−230)
For this A.P.,
First term, a = −5
nth term, an= −230
Common difference, d = a2−a1 =(−8)−(−5)
⇒d = − 8+5 = −3
Let −230 be the nth termof this A.P., and by the nth term formula we know,
an= a+(n−1)d
−230 = − 5+(n−1)(−3)
−225 = (n−1)(−3)
(n−1) = 75
n = 76
And, Sum of n term,
Sn = n/2 (a + l)
= 76/2 [(-5) + (-230)]
= 38(-235)
= -8930