**Nishant Verma**

Frederick Griffith.

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

**Question 1.** **Asked on :**01 September 2019:12:21:47 PM

Who discovered bacterial transformation?

*-Added by Akki chauhan* **Mathematics** » **Arithmetic Progressions**

**Answer:**

Frederick Griffith.

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

**Question 2.** **Asked on :**01 September 2019:12:20:02 PM

Which term is 78 of A.P 3,8,13,18...................

*-Added by Akki chauhan* **Mathematics** » **Arithmetic Progressions **

**Answer:**

∴

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

**Question 3.** **Asked on :**24 August 2019:09:16:21 PM

Find the sum of the first 25 multiples of 7.

*-Added by Prince Rawat* **Mathematics** » **Arithmetic Progressions **

**Answer:**

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}

S_{25} = 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

**Question 4.** **Asked on :**24 August 2019:09:15:05 PM

Find the sum of n terms of A.P. 2+4+6+...... .

*-Added by Prince Rawat* **Mathematics** » **Arithmetic Progressions **

**Answer:**

Answer:30,240

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

**Question 5.** **Asked on :**24 August 2019:09:11:37 PM

Find the sum of the A.p.1+3+5+7+.....+199.

*-Added by Prince Rawat* **Mathematics** » **Arithmetic Progressions **

**Answer:**

an = a+ (n-1)d

199 = 1 + (n-1)2

198 = (n-1)2

198/2 = n - 1

99 = n - 1

n = 100

No: of terms = n = 100

Sum is 100.

Sum = Sn

Sn = n/2 [a + an]

= 100/2 [ 1 + 199]

= 50 [ 200]

= 10000

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

**Question 6.** **Asked on :**17 August 2019:09:27:38 AM

What is the form of Arithmetic Progressions?

*-Added by Akki chauhan* **Mathematics** » **Arithmetic Progressions **

**Answer:**

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

**Question 7.** **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:**

Given: a_{1} = 6, S_{6 }= 66

using Formula: S_{n} = n2 (*a +* *l*)

**Step by Step Explanation:**

S_{6} = 62 ( 6 + a_{6})

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

**Question 8.** **Asked on :**09 July 2019:07:05:00 PM

If the sum of n^{th} terms of an A.P. is given by 4n^{2} - 7n. Find

(i) Sum of 23 terms

(ii) 15^{th} term

(iii) n^{th} term

*-Added by ATP Admin* **Mathematics** » **Arithmetic Progressions **

**Answer:**

**Given**:S_{n} = 4n^{2} - 7n

**Step by step** explanation:

**(i) Sum of 23 terms **

S_{n} = 4n^{2} - 7n ............. (**i**)

Replace n by 23 we have,

S_{23} = 4(23)^{2} - 7(23)

= 4 × 529 - 161

= 2116 - 161

= 1955 **Answer**

**(ii) 15 ^{th} term**

a_{15} = S_{15} - S_{14}

S_{15} = 4(15)^{2} - 7(15) = 4 × 225 - 105 = 900 - 105 = 795

S_{14} = 4(14)^{2} - 7(14) = 4 × 196 - 98 = 784 - 98 = 686

Now, a_{15} = S_{15} - S_{14}

= 795 - 686

= 109 **Answer**

(iii) n^{th} term

a_{n} = S_{n} - S_{n-1}

Given, S_{n} = 4n^{2} - 7n

Replace n by n-1 we have

S_{n-1} = 4(n-1)^{2} - 7(n-1)

= 4(n^{2} - 2n + 1) - 7n + 7

= 4n^{2} - 8n + 4 - 7n + 7

= 4n^{2} - 15n + 11 ............. (ii)

Now Applying equation (i) and (ii) in formula a_{n} = S_{n} - S_{n-1}

a_{n} = (4n^{2} - 7n) - (4n^{2} - 15n + 11)

= 4n^{2} - 7n - 4n^{2} + 15n - 11

= 8n - 11

Therefore, a_{n} = 8n - 11 **Answer**

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

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

**Question 9.** **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 Harshita Rathore* On 26 August 2019:03:25:54 PM

**Question 10.** **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 Master Purushottam* On 22 August 2019:08:54:23 PM

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