(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 = a2 − a1 = 7 − 2 = 5
n = 10
We know that the formula for the sum of the nth term in the AP series is,
Sn = n/2 [2a + (n – 1)d]
S10 = 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 = a2− a1
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,
Sn = n/2 [2a + (n – 1)d]
S12 = 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 = a2 − a1 = 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,
Sn = n/2[2a + (n – 1)d]
S12 = 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 = a2 – a1 = (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,
Sn = n /2 [2a + ( n – 1) d ]
Sn = (11/2) [2(1/15) + (11 – 1) × 1/ 60]
= 11/2(2/15 + 10/60)
= 11/2 (9/30)
= 33/20