Question

How many terms of AP 9,17,25,.... mist be taken to give a sum of 636

Answer 1

Hey sup!

As per the question,

9,17,25......

a=9.
d=17-9=8.
S(n)=636.

S(n)=n/2(2a+(n-1)d).
=>636=n/2(2*9+(n-1)8).
=>636=n/2(18+8n-8).
=>636*2=n(18+8n-8).
=>1272=n(10+8n).
=>1272=10n+8n^2.
=>8n^2+10n-1272=0.
=>2(4n^2+5n-636)=0
=>4n^2+5n-636=0/2=0.
=>4n^2-48n+53n-636=0.
=>4n(n-12)+53(n-12)=0.
=>(4n+53)(n-12)=0

We'll discard (4n+53) as it gives -ve value.
n-12=0.
n=12.

So sum of 12 term is 636.

Hope it helps.

Answer 2

Solution:

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Given:

AP: 9,17,25...

∴ a = 9,

∴ $d=a_2 -a_1 = 17 - 9 = 8$

Sum of terms = 636,

$S_n = 636$

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To Find:

Number of terms,to from sum 636,.

n = ?

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We know that,

$S_n = 636,$

&

$S_n = \frac{n}{2} (2a+(n-1)d)$

=> $636 = \frac{n}{2} (2(9)+(n-1)8)$

=> $636 = \frac{n}{2} (18+8n-8)$

=> $636 = \frac{n}{2} (8n+10)$

=> $636 = n(4n+5)$

=> $636 = 4n^2 +5n$

=> $4n^2 +5n -636 =0$

=> $4n^2 -48n + 53n -636 = 0$

=> $4n(n - 12) + 53 (n - 12) = 0$

=> $(4n + 53) (n-12) = 0$

We know that,

anything x 0 = 0,.

So,

4n- 53 = 0                                 (or)  n -12 =0
4n = 53                                      (or)  n = 12,
  n = $\frac{53}{4}$      (or)  n = 12,.. (number of terms can't be in fraction)

                   ∴ n = 12,.

               ∴ The number of terms whose sum is 636 is 12,

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                                     Hope it Helps!!