Wellcome!

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

#### Who discovered bacterial transformation?

-Added by Akki chauhan Mathematics » Arithmetic Progressions

Nishant Verma
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

Nishant Verma
78 is the 16th term of the given A.P. 16th term of A.P : 3,8,13,18,is 78.

-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

Master Mind

The A.P. : 7, 14, 21, ..............

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

∴ a = 7, d = 7 n = 25

Sn = n2 {2a + (n - 1)d}

S25 = 252 {2 × 7 + (25 - 1)7}

= 252 {14 + (24)7}

= 252 (14 + 168)

= 252 × 182

= 25 × 91

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

Akki chauhan

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

Akki chauhan
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

Himanshi Verma

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.

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

-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 nth terms of an A.P. is given by 4n2 - 7n. Find (i) Sum of 23 terms (ii) 15th term(iii) nth term

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

(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

(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 9. 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.

Harshita Rathore

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

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

Master Purushottam

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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