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Asked/Added Questions From NCERT:


Question 1. Asked on :10 July 2019:11:49:11 PM

The first term of an A.P. is 6. The sum of first 6 terms is 66. Find it's 6th term.

-Added by ATP Admin Mathematics » Arithmetic Progressions

Answer:

Master Mind

Given: a1 = 6, S= 66 

using Formula: Sn = n2 (a + l

Step by Step Explanation: 

S6 = 62 ( 6 + a6)

66 = 3(6 + 6th term) 

(6 + 6th term) = 663 = 22

6th term = 22 - 6 

6th term = 16 Answer 

-Answered by Master Mind On 11 July 2019:09:08:06 PM


Question 2. Asked on :09 July 2019:07:05:00 PM

If the sum of nth terms of an A.P. is given by 4n2 - 7n. Find 

(i) Sum of 23 terms 

(ii) 15th term

(iii) nth term

-Added by ATP Admin Mathematics » Arithmetic Progressions

Answer:

Master Mind

Given:Sn = 4n2 - 7n 

Step by step explanation: 

(i) Sum of 23 terms 

Sn = 4n2 - 7n ............. (i

Replace n by 23 we have,

S23 = 4(23)2 - 7(23) 

= 4 × 529 - 161

= 2116 - 161 

= 1955 Answer

(ii) 15th term 

a15 = S15 - S14 

S15 = 4(15)2 - 7(15) = 4 × 225 - 105 = 900 - 105 = 795 

S14 = 4(14)2 - 7(14) = 4 × 196 - 98 = 784 - 98 = 686

Now, a15 = S15 - S14 

              = 795 - 686 

              = 109 Answer

(iii) nth term 

an = Sn - Sn-1

Given, Sn = 4n2 - 7n 

Replace n by n-1 we have 

Sn-1 = 4(n-1)2 - 7(n-1) 

       = 4(n2 - 2n + 1) - 7n + 7

       = 4n2 - 8n + 4 - 7n + 7 

       = 4n2 - 15n + 11 ............. (ii) 

Now Applying equation (i) and (ii) in formula an = Sn - Sn-1 

an = (4n2 - 7n) - (4n2 - 15n + 11) 

    = 4n2 - 7n - 4n2 + 15n - 11

    = 8n - 11 

Therefore, an = 8n - 11 Answer  

*******************************

-Answered by Master Mind On 11 July 2019:09:42:19 PM


Question 3. 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 4. 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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