Asked/Added Questions From NCERT:


Question 1. Asked on :29 December 2018:01:00:20 AM

If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.

-Added by ATP Admin Mathematics » Arithmetic Progressions

Answer:

Himanshi Verma

  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

How many terms of the AP : 9, 17, 25, . . . must be taken to give a sum of 636?

-Added by ATP Admin Mathematics » Arithmetic Progressions

Answer:

Himanshi Verma

 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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