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Class 10 > Solve the equation: 1 + 4 + 7 + 10 + ... + x = 287
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Solve the equation: 1 + 4 + 7 + 10 + ... + x = 287

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Physicist

X=40 First use the formula of summation of n terms an A.P and get value of n then use formula of nth term of A.P
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airthmetic progression Sn = n/2(2a+(n-1)d) given a=1, d=4-1=3 & Sn = 287 287 = n/2 (2*1 +(n-1) 3) 287*2 = n(2 + 3n - 3) 574 = 2n + 3n^2 - 3n 3n^2 -n - 574 = 0 on solving the quadratic equation using formula n= -b + sq.root(b^2 -4ac) ...
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airthmetic progression Sn = n/2(2a+(n-1)d) given a=1, d=4-1=3 & Sn = 287 287 = n/2 (2*1 +(n-1) 3) 287*2 = n(2 + 3n - 3) 574 = 2n + 3n^2 - 3n 3n^2 -n - 574 = 0 on solving the quadratic equation using formula n= -b + sq.root(b^2 -4ac) ----------------------- 2a we get n = 14, -41/3 n not equal to -41/3 due to negative nos. n=14 Sn = n/2 (a +l) 287 = 14/2(1 +x) 574 = 14 (1+x) 574 / 14 = 1+x 41 = 1 + x So, x = 41 - 1 x = 40 is the solution read less
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Solution is: x = 40 Explanation:- The LHS of the given equation is an AP, Let the total no.of terms upto 'x' are N, So now using the suitable formula for sum of AP; (N/2) = S Here, a = 1, d = 3, N = ?, S = 287 So, on putting all these values and solving, we get; N = 14 or N = -82/6 Accepting, N =...
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Solution is: x = 40 Explanation:- The LHS of the given equation is an AP, Let the total no.of terms upto 'x' are N, So now using the suitable formula for sum of AP; (N/2)[ 2a + (N-1)d ] = S Here, a = 1, d = 3, N = ?, S = 287 So, on putting all these values and solving, we get; N = 14 or N = -82/6 Accepting, N = 14 (being a natural number) Further to solve the value of 'x' , using the formula for finding Nth term of an AP; T = a + (N-1)d Here, T= x, N = 14, Therefore, on solving, we get; x = 40 read less
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