If an A.P. has a = 1, tn = 20 and sn = 399, then value of n is :
A. 20
B. 32
C. 38
D. 40
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & {S_n} = \frac{1}{2}\left( {a + l} \right) \times n \cr & \Rightarrow 399 = \left( {1 + 20} \right) \times \frac{n}{2} \cr & \Rightarrow 399 \times 2 = 21 \times n \cr & \Rightarrow n = 399 \times \frac{2}{{21}} \cr & \Rightarrow n = 19 \times 2 \cr & \Rightarrow n = 38 \cr} $$Related Questions on Progressions
Find the first term of an AP whose 8th and 12th terms are respectively 39 and 59.
A. 5
B. 6
C. 4
D. 3
E. 7
The sum of the first 16 terms of an AP whose first term and third term are 5 and 15 respectively is
A. 600
B. 765
C. 640
D. 680
E. 690
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