(1) + (1 + 1) + (1 + 1 + 1) + ....... + (1 + 1 + 1 + ...... n - 1 times) = ......
A. $$\frac{{n\left( {n + 1} \right)}}{2}$$
B. $$\frac{{n\left( {n - 1} \right)}}{2}$$
C. $${n^2}$$
D. $$n$$
Answer: Option B
Solution(By Examveda Team)
$$\eqalign{ & {S_n} = \left[ {2a + \left( {n - 1} \right)d} \right] \times \frac{n}{2} \cr & {\text{Here,}} \cr & d = 1 \cr & a = 1\, \cr & {\text{and}}\,\,n - 1\,\,{\text{terms}} \cr} $$$$ \Rightarrow {S_{n - 1}} = \left[ {2 + \left( {n - 1 - 1} \right)} \right]$$ $$ \times \frac{{\left( {n - 1} \right)}}{2}$$
$$ \Rightarrow {S_{n - 1}} = \left[ {2 + n - 2} \right]$$ $$ \times \frac{{\left( {n - 1} \right)}}{2}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{n\left( {n - 1} \right)}}{2}$$
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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