The number of terms of the A.P. 3, 7, 11, 15, ... to be taken so that the sum is 406 is

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In the given problem, we have an A.P. 3,7,11,15,.... Here, we need to find the number of terms n such that the sum of n terms is 406. So here, we will use the formula, S_n = n/2[2a + (n-1)d ] Where; a = first term for the given A.P. d = common difference of the given A.P. n = number of terms The first term (a) = 3 The sum of n terms (Sn) = 406 Common difference of the A.P. (d) = a_2 - a_1 = 7 - 3 = 4 So, on substituting the values in the formula for the sum of n terms of an A.P., we get, `406 = n/2 [ 2(3) + (n-1) (4) ] 406 = (n/2) [ 6 +(4n- 4)] 406 = (n/2) [ 2 + 4n] 406 = n + 2n^2` So, we get the following quadratic equation, 2n^2 + n - 406 = 0 On solving by splitting the middle term, we get, 2n^2 - 28n + 29n - 406 = 0 2n ( n- 14) - 29 (n-14)= 0 (2n - 29 ) ( n- 14) = 0 Further, 2n - 29 = 0 n = $\frac{29}{2}$ Or, n - 14 = 0 n = 14 Since, the number of terms cannot be a fraction, the number of terms (n) is n = 14

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