Asked/Added Questions From NCERT:

-Answered by Nishant Verma On 02 September 2019:11:45:39 AM

-Answered by Nishant Verma On 02 September 2019:10:23:45 AM

The A.P. : 7, 14, 21, .............. 

Now We have to find 7 + 14 + 21 + ........... 25 terms

∴ a = 7, d = 7 n = 25 

S_n = n2 {2a + (n - 1)d}

S₂₅ = 252 {2 × 7 + (25 - 1)7}

      = 252 {14 + (24)7}

      = 252 (14 + 168)

      = 252 × 182

      = 25 × 91 

      = 2275 Answer

Hence the sum of first 25 multiple of 7 is 2275.

-Answered by Master Mind On 24 August 2019:11:38:04 PM

Step-by-step explanation:

Let a be the first term and d the common difference

          A=2

        D=4-2=2

       So, 15th term=a+(15-1)*d

       =2+(14*2)

       2+28

       =30

Sum of first 15 term=

n/2(a+an),where n=number

of terms. 

=15/2(2+30)

=15/2(32)

=15*16

=240

-Answered by Akki chauhan On 24 August 2019:09:27:13 PM

-Answered by Akki chauhan On 27 August 2019:09:29:28 AM

 The general form of an Arithmetic progression is a, a+1, a+2, a+3.....

-Answered by Himanshi Verma On 17 August 2019:10:41:29 AM

Given: a₁ = 6, S₆ = 66 

using Formula: S_n = n2 (a + l) 

Step by Step Explanation: 

S₆ = 62 ( 6 + a₆)

66 = 3(6 + 6^th term) 

(6 + 6^th term) = 663 = 22

6^th term = 22 - 6 

6^th term = 16 Answer 

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

Given:S_n = 4n² - 7n 

Step by step explanation: 

(i) Sum of 23 terms 

S_n = 4n² - 7n ............. (i) 

Replace n by 23 we have,

S₂₃ = 4(23)² - 7(23) 

= 4 × 529 - 161

= 2116 - 161 

= 1955 Answer

(ii) 15^th 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) n^th term 

a_n = S_n - S_n-1

Given, S_n = 4n² - 7n 

Replace n by n-1 we have 

S_n-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 a_n = S_n - S_n-1 

a_n = (4n² - 7n) - (4n² - 15n + 11) 

    = 4n² - 7n - 4n² + 15n - 11

    = 8n - 11 

Therefore, a_n = 8n - 11 Answer  

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

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 Harshita Rathore On 26 August 2019:03:25:54 PM

 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 Master Purushottam On 22 August 2019:08:54:23 PM