Question

How many terms of an A. P. 1,4,7, _ _ _ _ _ _ are needed to give the sum 2380?

Answer 1

Here the terms are 1 , 4 , 7 ....
So a = 1 and common difference d = 4-1 = 7-4 = 3
And sum of AP = Sn = n/2 [2a + (n-1)d]
As Sn = 2380

Hence 2380 = n/2 [ 2*1 + (n - 1) 3 ]
4760 = n[2 + 3n -3 ]
4760 = 2n +3n2 -3n
4760 = 3n2 -n
Or 3n2 -n -4760 = 0
So n = 40 and n = -119/3

Number of terms can never be negative , so n = 40

Answer 2

we have a=1,d=3,sn=2380
so ,n= no. of terms
formula, sn=n/2 (2a+(n-1)d)
2380= n/2(2*1+(n-1)3)
4760= 2n +3 n2-3n
solve ,3n2-n-4760