how many terms of AP 9,17,25..must be taken to give sum636
Answers
Step-by-step explanation:
Let the number of terms required to make the sum of 636 be n and common difference be d.
Given Arithmetic Progression : 9 , 17 , 25 ....
First term = a = 9
Second term = a + d = 17
Common difference = d = a + d - a = 17 - 9 = 8
From the indentities of arithmetic progressions, we know : -
S_{n}=\dfrac{n}{2}\{2a+(n-1)d\}S
n
=
2
n
{2a+(n−1)d} , where S_{n}S
n
is the sum of first term to nth term of the AP, a is the first term, d is the common difference and n is the number of terms of AP.
In the given Question, sum of APs is 636.
Therefore,
\begin{gathered} = > 636 = \dfrac{n}{2} \{2(9) + (n - 1)8 \} \\ \\ \\ = > 1272 = n(18 + 8n - 8) \\ \\ = > 1272 = 10n + 8n {}^{2} \\ \\ = > 636 = 5n + 4n {}^{2} \end{gathered}
=>636=
2
n
{2(9)+(n−1)8}
=>1272=n(18+8n−8)
=>1272=10n+8n
2
=>636=5n+4n
2
= > 4n² + 5n - 636 = 0
= > 4n² + ( 53 - 48 )n - 636 = 0
= > 4n² + 53n - 48n - 636 = 0
= > 4n² - 48n + 53n - 636 = 0
= > 4n( n - 12 ) + 53( n - 12 ) = 0
= > ( n - 12 )( 4n + 53 ) = 0
By Zero Product Rule,
= > n - 12 = 0
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= > n = 12
a1=9
a2=a1+d=17
9+d=17
d=8
We know that, the sum of n terms of an A.P. is,
Sn=2n{2a+(n−1)d}
636=2n{2(9)+(n−1)8}
1272=18n+8n2−8n
8n2+10n−1272=0
4n2+5n−636=0
n=12,−13.25
Ignore the negative value.
n=12
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