Question

How many terms of the A.P. 9,17,25,-------must be taken to give a sums of 636?

Answer 1

Answer:

given series ,

9,17,25,....is in Ap with common difference 8.

we know that,

Sn= n/2[2a+(n-1)d],

=> 636 = n/2[2*9+(n-1)8],

=> 1272 = 18n + 8n^2-8n,

=> 8n^2+10n-1272 =0,

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

=> n = [-5+/-√(25-4*4*(-636))]/(2*4),

=> n =[-5+/-(101)]/8,

=> n = 96/8 or -106/8,

=> n = 12.

therefore number of terms = 12.

Answer 2

$\large\sf\underline\red{GIVEN:-}$

$\large\tt\green{a=9,}$

$\large\tt\green{d=17-9=8}$

$\large\tt\green{Sn=636}$

$\small\sf\underline\pink{The\:sum\:of\:n\:terms\:of\:an\:AP\:is\:given\:by}$

$\longrightarrow$$\large\tt\purple{Sn=\frac{n}{2}(2a+(n-1)d)}$

$\longrightarrow$$\large\tt\purple{636=\frac{n}{2}(2(9)+(n-1)(8))}$

$\longrightarrow$$\large\tt\purple{636=n(9+4n-4)}$

$\longrightarrow$$\large\tt\purple{{4n}^{2} + 5n - 636 = 0}$

$\longrightarrow$$\large\tt\purple{{4n}^{2} - 53n - 48n - 636 = 0}$

$\longrightarrow$$\large\tt\purple{n(4n+53)-12(4n+53)=0}$

$\longrightarrow$$\large\tt\purple{(n-12)(4n+53)=0}$

$\longrightarrow$$\large\tt\purple{n-12=0(\because 4n+53≠0\:as\:n≠\frac{-53}{4}}$

$\longrightarrow$$\large\tt\purple{n=12}$