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 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:
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 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:
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
-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 Himanshi Verma On 23 March 2019:05:36:28 PM
Question 4. 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 Himanshi Verma On 20 January 2019:05:32:22 PM
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