The best way to solve this hard question of chapter arithmetic progressions of exercise 5.3 of math . What is the easy solution In an AP(v) Given d = 5, S9 = 75, find a and a9.
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Given that, d = 5, S9 = 75
As, sum of n terms in AP is,
Sn = n/2 [2a +(n -1)d]
Therefore, the sum of first nine terms are;
S9 = 9/2 [2a +(9-1)5]
25 = 3(a+20)
25 = 3a+60
3a = 25−60
a = -35/3
As we know, the nth term can be written as;
an = a+(n−1)d
a9 = a+(9−1)(5)
= -35/3+8(5)
= -35/3+40
= (35+120/3) = 85/3