This is the Important question based on Arithmetic progression Chapter of R.S Aggarwal book for ICSE & CBSE Board.
In this Question the first term, its nth term and the sum to first nth terms is given
you have to find the n and the common difference.
Question Number 30 Of Exercise 11 C of RS Aggarwal Solution.
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In an AP the first term is 22, nth term is -11 and sum to first nth terms is 66. Find n and d , the common difference. Hint= a=22 , l= -11 and n/2 (a+l) = 66
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Given, first term a = 22.
Given, nth term[(a + n – 1) * d] = -11. ——— (1)
Now,
We know that Sum of n terms of an AP Sn = (n/2)[2a + (n – 1) * d]
⇒ 66 = (n/2)[a + a + (n – 1) * d]
⇒ 132 = n[22 – 11]
⇒ 132 = n[11]
⇒ n = 12.
Now,
We know that nth term of an AP is an = a + (n – 1) * d
⇒ -11 = 22 + (12 – 1) * d
⇒ -11 = 22 + 11 * d
⇒ -33 = 11d
⇒ d = -3.