Let 1st term of AP be a
Its common difference
d = 3rd term - 2nd term = 18-14=4
So a+d =14=> a=14-4=10
Sum of first 51 terms =51/2(2*10+(51-1)*4)
= 51×110=5610
Sum of first n terms of A.P will be n2
Explanation :
n1=7
Sn=49
we know that Sn = n/2{2a+(n-1)d}
thus,
Sn1=7/2(2a+6d)=49
=14a+42d=98 ... (1) x17
n2=17
Sn2=289
Sn2=17/2(2a+16d)=289
=34a+272d=578 ....(2) x7
from eq 1 & 2
238a +714d =1666 ...
Central Board of Secondary Education, CBSE concluded the Compartment Exams 2017 for Classes 10th and 12th on July 24. Students who did not qualify the CBSE 10th and 12th Exams...
(d) Of the 16 Mahajanapadas, Magadha, Kosala, Vatsa and Avanti were more powerful. They fought amongst themselves for years and ultimately Magadha emerged victorious under Bimbisara (Haranyak dynasty) in 6th...