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 Solution For arithemetic progression f( 7 ) = 13 , S( 14 ) = 203 , then find f( 10 ) and S( 8 ).Solution:Your problem -> For arithemetic progression f( 7 ) = 13 , S( 14 ) = 203 , then find f( 10 ) and S( 8 ).We have given f(7) = 13, S_14 = 203 and we have to find f(10) = ? and S(8) = ?We know that, f(n) = a + (n - 1)df(7) = 13=> a + (7 - 1)d = 13=> a + 6d = 13 ->(1)We know that, S_n = n/2 [2a + (n - 1)d]S_14 = 203=> 14/2 * [2a + (14 - 1)d] = 203=> 7 * [2a + 13d] = 203=> 2a + 13d = 29 ->(2)Solving (1) and (2), we get a = -5 and d = 3We know that, f(n) = a + (n - 1)df(10) = -5 + (10 - 1)(3)= -5 + (27)= 22We know that, S_n = n/2 [2a + (n - 1)d]:. S_8 = 8/2 * [2(-5) + (8 - 1)(3)]= 4 * [-10 + (21)]= 4 * [11]= 44 Solution provided by AtoZmath.com Any wrong solution, solution improvement, feedback then Submit Here Want to know about AtoZmath.com and me