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:
-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
Sn = 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
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:
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:
-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: a1 = 6, S6 = 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 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
-Added by ATP Admin Mathematics » Arithmetic Progressions
Answer:
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 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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