Question

If the nth term of A.p. is ( 2n + 1 ) then the sum of its and term is....
Key points :- n(n-2) , n( n+2) , n(n+1) , n squre

Answer 1

Given, a

Given, a n

Given, a n

Given, a n =2n−1

Given, a n =2n−1if n=1, then

Given, a n =2n−1if n=1, thena

Given, a n =2n−1if n=1, thena 1

Given, a n =2n−1if n=1, thena 1

Given, a n =2n−1if n=1, thena 1 =2(1)−1=1

Given, a n =2n−1if n=1, thena 1 =2(1)−1=1S

Given, a n =2n−1if n=1, thena 1 =2(1)−1=1S n

Given, a n =2n−1if n=1, thena 1 =2(1)−1=1S n

Given, a n =2n−1if n=1, thena 1 =2(1)−1=1S n =

Given, a n =2n−1if n=1, thena 1 =2(1)−1=1S n = 2

Given, a n =2n−1if n=1, thena 1 =2(1)−1=1S n = 2n

Given, a n =2n−1if n=1, thena 1 =2(1)−1=1S n = 2n

Given, a n =2n−1if n=1, thena 1 =2(1)−1=1S n = 2n (a

Given, a n =2n−1if n=1, thena 1 =2(1)−1=1S n = 2n (a 1

Given, a n =2n−1if n=1, thena 1 =2(1)−1=1S n = 2n (a 1

Given, a n =2n−1if n=1, thena 1 =2(1)−1=1S n = 2n (a 1 +a

Given, a n =2n−1if n=1, thena 1 =2(1)−1=1S n = 2n (a 1 +a n

Given, a n =2n−1if n=1, thena 1 =2(1)−1=1S n = 2n (a 1 +a n

Given, a n =2n−1if n=1, thena 1 =2(1)−1=1S n = 2n (a 1 +a n )

Given, a n =2n−1if n=1, thena 1 =2(1)−1=1S n = 2n (a 1 +a n )⇒S

Given, a n =2n−1if n=1, thena 1 =2(1)−1=1S n = 2n (a 1 +a n )⇒S n

Given, a n =2n−1if n=1, thena 1 =2(1)−1=1S n = 2n (a 1 +a n )⇒S n

Given, a n =2n−1if n=1, thena 1 =2(1)−1=1S n = 2n (a 1 +a n )⇒S n =

Given, a n =2n−1if n=1, thena 1 =2(1)−1=1S n = 2n (a 1 +a n )⇒S n = 2

Given, a n =2n−1if n=1, thena 1 =2(1)−1=1S n = 2n (a 1 +a n )⇒S n = 2n

Given, a n =2n−1if n=1, thena 1 =2(1)−1=1S n = 2n (a 1 +a n )⇒S n = 2n

Given, a n =2n−1if n=1, thena 1 =2(1)−1=1S n = 2n (a 1 +a n )⇒S n = 2n (1+2n−1)

Given, a n =2n−1if n=1, thena 1 =2(1)−1=1S n = 2n (a 1 +a n )⇒S n = 2n (1+2n−1)⇒S

Given, a n =2n−1if n=1, thena 1 =2(1)−1=1S n = 2n (a 1 +a n )⇒S n = 2n (1+2n−1)⇒S n

Given, a n =2n−1if n=1, thena 1 =2(1)−1=1S n = 2n (a 1 +a n )⇒S n = 2n (1+2n−1)⇒S n

Given, a n =2n−1if n=1, thena 1 =2(1)−1=1S n = 2n (a 1 +a n )⇒S n = 2n (1+2n−1)⇒S n =n

Given, a n =2n−1if n=1, thena 1 =2(1)−1=1S n = 2n (a 1 +a n )⇒S n = 2n (1+2n−1)⇒S n =n 2