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Solve the equation : 1 + 4 + 7 + 10 +...+ x =287. Solution: Given, the sum of the terms up to x is 287. We have to solve the equation. The sum of the first n terms of an AP is given by. Sₙ = n/2[2a + (n-1)d] Here, first term, a = 1. Common difference, d = 4 - 1 = 3. So, 287 = n/2[2(1) + (n - 1)3] 287 = n/2[2 + 3n - 3] 287 = n/2[3n - 1] 574 ...
Given, 1 + 4 + 7 + 10..... + x = 287. We need to find the value of x, First-term, a = 1. Common difference, d = 4-1 = 3. S n = 287. Using the formula, S n = n 2 (2 a + (n − 1) d) 287 = n 2 (2 × 1 + (n-1) 3) 287 = n 2 (2 + 3 n-3) 574 = n (3 n-1) 574 = 3 n 2-n. 3 n 2-n-574 = 0. 3 n 2-42 n + 41 n-574 = 0. 3 n (n-14) + 41 (n-14) = 0 (3 n + 41 ...
26 sie 2020 · Here, 1, 4, 7, 10, ..., x form an AP with a = 1, d = 3, an = x. We have, an = a + (n – 1)d. So, x = 1 + (n – 1) × 3 = 3n – 2. Also, S = n/2 (a + l) So, 287 =n/2 (1+ x) As n cannot be negative, so n = 14. Therefore, x = 3n – 2 = 3 × 14 – 2 = 40.
Solve for x : 1 + 4 + 7 + 10 + … + x = 287. any term - preceding term = 3 = common difference. Let there be n terms so, x = a n. Given, a = 1, d = 3 and S n = 287. By formula, S n = \dfrac {n} {2} [2a + (n - 1)d] 2n[2a+ (n−1)d]
Find the sum to n term of the A.P. 5, 2, −1, −4, −7, ..., The first and last terms of an AP are a and l respectively. Show that the sum of the nth term from the beginning and the nth term form the end is ( a + l ). Choose the correct alternative answer for the following question . If for any A.P. d = 5 then t 18 – t 13 = ....
For example displaystyle (x-1) (x-2) and displaystylex²-3x+2 are two polynomial expressions that represent the same polynomial; so, one has the equality displaystyle (x-1) (x-2)=x²-3x+2.
