First term of A.P. = 9 , common difference = 8if sum of n terms 636, Sn = (n/2) [ 2×9 + (n-1)8 ] = 636Sn = n [ 10+ 8n] = 12728n2 +10n -1272 = 04n2 +5n -636 = 0By factorising LHS,(4n+53)(n-12) = 0hence n = 12
12
12
12 terms should be taken
12

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 = 9Second term = a + d = 17Common difference = d = a + d – a = 17 – 9 = 8From the indentities of arithmetic progressions, we know : -{tex}S_{n}=\\dfrac{n}{2}\\{2a+(n-1)d\\}, where S_{n} {/tex}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,\xa0⇒ {tex}636 = \\dfrac{n}{2} \\{2(9) + (n – 1)8 \\} \\\\ \\\\ \\\\ = > 1272 = n(18 + 8n – 8) \\\\ \\\\ = > 1272 = 10n + 8n {}^{2} \\\\ \\\\ = > 636 = 5n + 4n {}^{2}{/tex}\xa0⇒ 4n² + 5n – 636 = 0\xa0⇒ 4n² + ( 53 – 48 )n – 636 = 0\xa0⇒ 4n² + 53n – 48n – 636 = 0\xa0⇒ 4n² – 48n + 53n – 636 = 0\xa0⇒ 4n( n – 12 ) + 53( n – 12 ) = 0\xa0⇒ ( n – 12 )( 4n + 53 ) = 0By Zero Product Rule,\xa0⇒ n – 12 = 0\xa0⇒ n = 12Hence,Number of terms of the AP [ 9 , 17 , 25 ] which are required to make the sum of 636 is 12.\xa0

It\’s 12
Lamba hai

636 is the sum of 12 terms
12 is the answer Hi are u single

12

a=9a2=17 difference =a2-a1 =17-9=8sum=636 put the formula n get ans

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 = 9Second term = a + d = 17Common difference = d = a + d – a = 17 – 9 = 8From the indentities of arithmetic progressions, we know : -{tex}S_{n}=\\dfrac{n}{2}\\{2a+(n-1)d\\}, where S_{n}{/tex} 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,{tex}= > 636 = \\dfrac{n}{2} \\{2(9) + (n – 1)8 \\} \\\\ \\\\ \\\\ = > 1272 = n(18 + 8n – 8) \\\\ \\\\ = > 1272 = 10n + 8n {}^{2} \\\\ \\\\ = > 636 = 5n + 4n {}^{2}{/tex}= > 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 ) = 0By Zero Product Rule,= > n – 12 = 0= > n = 12Hence,Number of terms of the AP [ 9 , 17 , 25 ] which are required to make the sum of 636 is 12.
12 terms must be taken to give a sum of 636…

According to the question,we have,\xa0a=9. Therefore, common difference d =17-9=8let the required number of terms be n.Therefore, Sn=636{tex}\\Rightarrow{/tex}{tex}\\frac{n}{2}{/tex}[2a+(n-1)d]=636{tex}\\Rightarrow{/tex}{tex}\\frac{n}{2}{/tex}[2(9)+(n-1)8]=636{tex}\\Rightarrow{/tex}n[18+8n-8]=1272{tex}\\Rightarrow{/tex}8n2+10n-1272=0{tex}\\Rightarrow{/tex}4n2+5n-636=0{tex}\\Rightarrow{/tex}4n2+53n-48n-636=0{tex}\\Rightarrow{/tex}n(4n+53)-12(4n+53)=0{tex}\\Rightarrow{/tex}(4n+53)(n-12)=0{tex}\\Rightarrow{/tex}4n+53=0 or n-12=0{tex}\\Rightarrow{/tex}n={tex}\\frac{{ – 53}}{4}{/tex}\xa0or n=12Since number of terms cannot neither be negative nor fraction, n=12hence, the required number of terms is 12.

Use the formula Sn=n/2[2a+(n-1)×d] and you will get the value of n.