Question

Find the sum of first n terms of the following A.P. ​

Answer 1

Answer:

The answer is Option A....7n-2/2

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Answer 2

Answer:

Sum of n terms = $[\frac{7n-1}{2} ]$

Step-by-step explanation:

Given problem

Find sum of n terms  AP

$[4- \frac{1}{n} ]$ + $[4-\frac{2}{n} ]$ + $[ 4-\frac{3}{n} ]$ ..  [up to n terms]

⇒ given series is in AP

⇒ first term a = $[ 4-\frac{1}{n}]$  

⇒ common difference d = a₂ - a₁  

                                         = $[4- \frac{2}{n} ] - [4 - \frac{1}{n} ]$

                                         = $[ \frac{4n - 2}{n} ] - [ \frac{4n - 1}{n} ]$

                                         = $[\frac{4n - 2 - 4n +1 }{n}]$  = $-\frac{1}{n}$  

⇒ sum of n terms S = $\frac{n}{2}[ 2a +(n-1)d]$

                                 = $\frac{n}{2} [ 2(4-\frac{1}{n} )+ (n-1)(-\frac{1}{n})]$  

                                 = $\frac{n}{2} [8 - \frac{1}{n} + n(-\frac{1}{n} ) + \frac{1}{n} ]$  

                                 = $\frac{n}{2} [8- \frac{2}{n} - 1 + \frac{1}{n} ]$  

                                 = $\frac{n}{2} [7 - \frac{1}{n} ]$  

                                 = $\frac{n}{2} [ \frac{7n-1}{n}]$  = $[\frac{7n-1}{2} ]$  

⇒ option (b) is correct