Question

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Answer 1

Answer:

Hence, Number of terms of the AP [ 9 , 17 , 25 ] which are required to make the sum of 636 is 12

Step-by-step explanation:

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Answer 2

Answer:

$\bf\huge{Solution:-}$

In the AP

a = 9, d = 17 -9 =8

Now,

$\bf{Sn= \:n/2 \:{2a+(n-1)d}}$

$\bf{=>636=n/2{2×9+(n-1)8}}$

$\bf{=> \:1272= \:n{18+8n-8}}$

=>1272= n{10+8n)

$\bf{=>1272=10n+8n²}$

$\bf{=>1272=2(5n+4n²)}$

$\bf{=>1272/2=5n+4n²}$

$\bf{=> 636 = 5n4n²}$

$\bf{=> 0=4n² + 5n - 636}$

$\bf{=>0=4n²+5n - 636}$

$\bf{=>0=n(4n²+53)-12(4n+53)}$

$\bf{=>0=(n-12)(4n+53)}$

$\bf{Either,n-12=0 or 4n-53=0}$

$\bf{\:n=2 \:n=53/4(not possible)}[\tex]</p><p></p><p></p><p></p><p>[tex]\bf{Therefore,n=12}$

$\bf{Hence, \:no \:of \:term=12}$