Question4
How many terms of the A.P. 6, 11/2, 5 . are needed to give the sum 25?
AP is of the form 6, 11/2, 5 .
Here
First term = a = 6
Common difference = d = 11/2 ( 6)
= 11/2 + 6
= ( 11 + 12 )/2
= 1/2
& Sum of n terms = Sn = 25
We need to find n
We know that
Sn = n/2 [2a + (n 1)d]
Here, Sn = 25 , a = 6 , d =1/2
Putting values
25 = n/2 ["2( 6) + (n 1)" (1/2)]
25 2 = n [" 12 + " n/2 " " 1/2]
50 = n [n/2 " 12 " 1/2]
50 = n [n/2 " " ((12(2) + 1)/2)]
50 = n [n/2 " " ((24 + 1)/2)]
50 = n [n/2 25/2]
50 = n [(n 25)/2]
50 2 = n(n 25)
100 = n2 25n
0 = n2 25n + 100
n2 25n + 100 = 0
n2 20n 5n + 100 = 0
n (n 20) 5 (n 20) = 0
(n 5)(n 20) = 0
Thus, n 5 = 0 , n 20 = 0
So, n = 5 , n = 20
Hence, n = 5 , 20 both values gives the required sum.
Made by
Davneet Singh
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo
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