. Write the nth tern if the sum of n terms of an AP is 2n?
Answers
Step-by-step explanation:
⇒S
⇒S 1
⇒S 1
⇒S 1 =2(1)+3(1)
⇒S 1 =2(1)+3(1) 2
⇒S 1 =2(1)+3(1) 2 =5
⇒S 1 =2(1)+3(1) 2 =5S
⇒S 1 =2(1)+3(1) 2 =5S n
⇒S 1 =2(1)+3(1) 2 =5S n
⇒S 1 =2(1)+3(1) 2 =5S n =
⇒S 1 =2(1)+3(1) 2 =5S n = 2
⇒S 1 =2(1)+3(1) 2 =5S n = 2n
⇒S 1 =2(1)+3(1) 2 =5S n = 2n
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}S
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}S n
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}S n
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}S n =
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}S n = 2
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}S n = 2n
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}S n = 2n
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}S n = 2n {a+a+(n−1)d}
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}S n = 2n {a+a+(n−1)d}S
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}S n = 2n {a+a+(n−1)d}S n
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}S n = 2n {a+a+(n−1)d}S n
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}S n = 2n {a+a+(n−1)d}S n =
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}S n = 2n {a+a+(n−1)d}S n = 2
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}S n = 2n {a+a+(n−1)d}S n = 2n
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}S n = 2n {a+a+(n−1)d}S n = 2n
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}S n = 2n {a+a+(n−1)d}S n = 2n {a+a
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}S n = 2n {a+a+(n−1)d}S n = 2n {a+a n
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}S n = 2n {a+a+(n−1)d}S n = 2n {a+a n
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}S n = 2n {a+a+(n−1)d}S n = 2n {a+a n }
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}S n = 2n {a+a+(n−1)d}S n = 2n {a+a n }2n+3n
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}S n = 2n {a+a+(n−1)d}S n = 2n {a+a n }2n+3n 2
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}S n = 2n {a+a+(n−1)d}S n = 2n {a+a n }2n+3n 2 =
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}S n = 2n {a+a+(n−1)d}S n = 2n {a+a n }2n+3n 2 = 2
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}S n = 2n {a+a+(n−1)d}S n = 2n {a+a n }2n+3n 2 = 2n
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}S n = 2n {a+a+(n−1)d}S n = 2n {a+a n }2n+3n 2 = 2n
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}S n = 2n {a+a+(n−1)d}S n = 2n {a+a n }2n+3n 2 = 2n {5+a
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}S n = 2n {a+a+(n−1)d}S n = 2n {a+a n }2n+3n 2 = 2n {5+a n
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}S n = 2n {a+a+(n−1)d}S n = 2n {a+a n }2n+3n 2 = 2n {5+a n
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}S n = 2n {a+a+(n−1)d}S n = 2n {a+a n }2n+3n 2 = 2n {5+a n }
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}S n = 2n {a+a+(n−1)d}S n = 2n {a+a n }2n+3n 2 = 2n {5+a n }4+6n=5+a
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}S n = 2n {a+a+(n−1)d}S n = 2n {a+a n }2n+3n 2 = 2n {5+a n }4+6n=5+a n
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}S n = 2n {a+a+(n−1)d}S n = 2n {a+a n }2n+3n 2 = 2n {5+a n }4+6n=5+a n
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}S n = 2n {a+a+(n−1)d}S n = 2n {a+a n }2n+3n 2 = 2n {5+a n }4+6n=5+a n
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}S n = 2n {a+a+(n−1)d}S n = 2n {a+a n }2n+3n 2 = 2n {5+a n }4+6n=5+a n ⇒a
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}S n = 2n {a+a+(n−1)d}S n = 2n {a+a n }2n+3n 2 = 2n {5+a n }4+6n=5+a n ⇒a n
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}S n = 2n {a+a+(n−1)d}S n = 2n {a+a n }2n+3n 2 = 2n {5+a n }4+6n=5+a n ⇒a n
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}S n = 2n {a+a+(n−1)d}S n = 2n {a+a n }2n+3n 2 = 2n {5+a n }4+6n=5+a n ⇒a n =6n−1
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}S n = 2n {a+a+(n−1)d}S n = 2n {a+a n }2n+3n 2 = 2n {5+a n }4+6n=5+a n ⇒a n =6n−1⇒a
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}S n = 2n {a+a+(n−1)d}S n = 2n {a+a n }2n+3n 2 = 2n {5+a n }4+6n=5+a n ⇒a n =6n−1⇒a r
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}S n = 2n {a+a+(n−1)d}S n = 2n {a+a n }2n+3n 2 = 2n {5+a n }4+6n=5+a n ⇒a n =6n−1⇒a r
⇒S 1 =2(1)+3(1) 2 =5S n = 2n {2a+(n−1)d}S n = 2n {a+a+(n−1)d}S n = 2n {a+a n }2n+3n 2 = 2n {5+a n }4+6n=5+a n ⇒a n =6n−1⇒a r =6r−1
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