Sum of first 10 terms of an arithmetic sequence is 320 and sum of first 15 terms is 705
a) write algebra of sum of the sequence
b) write algebra of the sequence?
Answers
Tn = 6n - 1 , Sn = 3n² + 2n
Step-by-step explanation:
Let say AP is
a , a+ d , a+ 2d , ..............................................., a + (n-1)s
Sum on n terms
= (n/2)(a + a +(n-1)d)
= (n/2)(2a + (n-1)d)
Sum of 10 terms = (10/2)(2a + 9d) = 320
=> 2a + 9d = 64
Sum of 15 Terms = (15/2)(2a + 14d) = 705
=> 2a + 14d = 94
=> 5d = 30
=> d = 6
=> a = 5
AP is
5 , 11 , 16 , ........................
nth Term = 5 + (n-1)6
= 6n - 1
Tn = 6n - 1
Sum = (n/2)(2a + (n-1)d)
= (n/2)(10 + (n-1)6)
= (n/2)(10 + 6n - 6)
= n(3n + 2)
= 3n² + 2n
Sn = 3n² + 2n
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