Question 1. Asked on :29 December 2018:01:00:20 AM
-Added by ATP Admin Mathematics » Arithmetic Progressions
Answer:
S7 = 7/2(2a + 6d)
=> 49 = 7a + 21d
=> 7 = a + 3d ------------ (i)
S17 = 17/2 (2a + 16d)
=> 289 = 17a + 136
=> 17 = a + 8d -------------- (ii)
Substracting equation (i) from (ii)
we have ...
17 - 7 = a - a + 8d - 3d
10 = 5d
d = 10/5
d = 2
Putting the value in Equation (i)
7 = a + 3d
7 = a + 3 x 2
a = 7 - 6
a = 1
Sum of first n terms = n/2 [2a + (n - 1) d]
= n/2 [2x 1 + (n - 1)2]
= n/2 [2 + 2n - 2 ]
= n/2 (2n)
= nxn = n2
-Answered by Akki chauhan On 14 January 2019:10:14:29 AM
Question 2. Asked on :28 December 2018:11:00:50 PM
-Added by ATP Admin Mathematics » Arithmetic Progressions
Answer:
A.p= 9,17,25
a=9
d=17-9
=8
Sum of n terms=n÷2[2a+(n-1)d
636=n÷2[2(9)+(n-1)8]
636=n÷2[18+8n-8]
636=n÷2[10+8n]
1272=10n+8n^2
8n^2+10n-1272=0
2[4n^2+5n-636]=0
4n^2+5n-636=0
4n^2+53n-48n-636=0
n(4n+53)-12(4n+53)=0
(n-12)(4n+53)=0
n-12=0
n=12
12 terms must be given to sum of 636
-Answered by Himanshi Verma On 20 January 2019:05:32:22 PM
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