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 Himanshi Verma On 23 March 2019:05:36:28 PM
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
You can see here all the solutions of this question by various user for NCERT Solutions. We hope this try will help you in your study and performance.
This Solution may be usefull for your practice and CBSE Exams or All label exams of secondory examination. These solutions or answers are user based solution which may be or not may be by expert but you have to use this at your own understanding of your syllabus.