Find the sum of the following APs.
(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
(i) Given, 2, 7, 12 ,…, to 10 terms
first term, a = 2
And common difference, d = a2 − a1
= 7−2 = 5
n = 10
sesum of nth term in 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
For this A.P.,
first term, a = −37
And common difference, d = a2− a1
d= (−33)−(−37)
= − 33 + 37 = 4
n = 12
We know that, the formula for sum of nth term in 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
For this A.P.,
first term, a = 0.6
Common difference, d = a2 − a1 = 1.7 − 0.6 = 1.1
n = 100
We know that, the formula for sum of nth term in 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
For this A.P.,
First term, a = 1/5
Common difference, d = a2 –a1
= (1/12)-(1/5) = 1/60
And number of terms n = 11
sum of nth term in AP series is,
Sn = n/2 [2a + (n – 1) d]
\(S_n = \frac{11}{2}[2(\frac{1}{15}) + \frac{(11 -1)}1{60}]\)
= 11/2(2/15 + 10/60)
= 11/2 (9/30)
= 33/20