The best solution for solving the problem in arithmetic progressions of exercise 5.3 of math, what is the best trick for solving this question in easy way In an APiii) Given a12 = 37, d = 3, find a and S12.
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Given that, a12 = 37, d = 3
As we know, from the formula of the nth term in an AP,
an = a+(n −1)d,
Therefore, putting the given values, we get,
⇒ a12 = a+(12−1)3
⇒ 37 = a+33
⇒ a = 4
Now, sum of nth term,
Sn = n/2 (a+an)
Sn = 12/2 (4+37)
= 246