Multiple choice

Consider an arithmetic progression with n terms. If the common difference is increased by 1, then nth term increases by 19. If the 5th term of the progression is 28 and the average of the first and last terms is 61, then the 10th term of the progression is:

  1. 54

  2. 56

  3. 58

  4. 60
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let terms be a, a+d, ..., a+(n-1)d. n-th term = a+(n-1)d. If d increases by 1, n-th term = a+(n-1)(d+1) = a+(n-1)d + (n-1). Given increase is 19, so n-1 = 19 => n = 20. 5th term = a+4d = 28. Average of first and last = (a + a+19d)/2 = a+9.5d = 61. System: a+4d=28, a+9.5d=61. Subtract: 5.5d = 33 => d = 6. Then a = 28 - 24 = 4. 10th term = a+9d = 4 + 9(6) = 4 + 54 = 58.

AI explanation

Let the arithmetic progression have a first term a, a common difference d, and n terms. The nth term is given by Tn = a + (n-1)d; increasing the common difference by 1 makes the nth term increase by (n-1), so n - 1 = 19 and n = 20. The average of the first and last terms equals the average of the progression, which is 61. Since the first and last terms average to 61, their sum is 122, giving the equation a + a + 19d = 122, or 2a + 19d = 122. The 5th term is T5 = a + 4d = 28; multiplying this by 4 gives 4a + 16d = 112. Multiplying the first equation by 2 gives 4a + 38d = 244; subtracting the previous result leaves 22d = 132, meaning d = 6. Solving for a gives a + 24 = 28, so a = 4. The 10th term is T10 = a + 9d = 4 + 9(6) = 58.

Keep practicing — related questions