wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Find the sum of the following series:
(i) 2 + 5 + 8 + ... + 182
(ii) 101 + 99 + 97 + ... + 47
(iii) (a − b)2 + (a2 + b2) + (a + b)2 + ... + [(a + b)2 + 6ab]

Open in App
Solution

(i) 2 + 5 + 8 + ... + 182
Here, the series is an A.P. where we have the following:
a = 2d=5-2 = 3an=1822+(n-1)(3)=1822+3n-3=1823n-1=1823n=183n=61 Sn =n22a+(n-1)dS61= 6122×2+61-1×3 =6122×2+60×3 = 5612

(ii) 101 + 99 + 97 + ... + 47
Here, the series is an A.P. where we have the following:
a = 101d=99-101 = -2an=47101+(n-1)(-2)=47101-2n+2=472n-2=542n=56n=28Sn =n22a+(n-1)d S28 = 2822×101+28-1×(-2) = 2822×101+27×(-2) = 2072

(iii) (a − b)2 + (a2 + b2) + (a + b)2 + ... + [(a + b)2 + 6ab]
Here, the series is an A.P. where we have the following:
a = (a-b)2d=a2+b2- (a-b)2 = 2aban=[(a + b)2+6ab](a-b)2+(n-1)(2ab)= [(a + b)2+6ab] a2+b2-2ab+2abn-2ab=[a2+b2+2ab+6ab] a2+b2-4ab+2abn=a2+b2+8ab 2abn=12ab n=6 Sn =n22a+(n-1)dS6= 622(a-b)2+6-1 2ab = 32(a2+b2-2ab)+10ab = 32a2+2b2-4ab+10ab = 32a2+2b2+6ab = 6a2+b2+3ab

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
The Number Line
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon