An arithmetic sequence of 10 positive integers has a fourth

Question

An arithmetic sequence consists of 10 positive integers. The fourth term of the sequence is 30, and another term of the sequence is 42. What is the sum of all possible values of the first term?

A. 75
B. 99
C. 105
D. 117
E. 165

pappal · Answer

given 4th term=a+3d=30; and another term is 42.
so probable values of d could be d=2,3,4,6 resulting in a=24,21,18,12
also one case might be where a=42 and d=-4 (neglecting the cases of any higher values of a to make the 4th term as 30, since value of d will be large enough to accommodate for next 6 integers to be positive; also if values of d are higher than d=-4 (such as -3,-2) then 4th term wouldn't be 30 but more)
sum of all values of a=24+21+18+12+42=117
D

achyut123 · Answer

Hi, Please provide the solution to this.

percyboi66 · Answer

I was able to get D by checking a + (n - 1)d = 42 for all values of n from 5 to 10 (+1 in case the first term, a is 42). However, I'm wondering if there is a quicker solution to this?

kevincan · Answer

An arithmetic sequence of 10 positive integers has first term a and common difference d. The terms are a, a + d, a + 2d, ..., a + 9d.

The fourth term is 30, so a + 3d = 30. This gives a = 30 − 3d.

Another term equals 42, so for some integer k between 0 and 9, a + kd = 42.

Substitute a = 30 − 3d into this equation:
(30 − 3d) + kd = 42
30 + (k − 3)d = 42
(k − 3)d = 12

Since k is an integer, k − 3 must be a divisor of 12. The possible values are ±1, ±2, ±3, ±4, ±6, ±12.

This is a crucial point: d need not be positive!

This gives valid k values of 4, 5, 6, 7, 9 for positive d, and 2, 1, 0 for negative d.

Now compute a = 30 − 3d.

For positive d:
d = 6 → a = 12
d = 4 → a = 18
d = 3 → a = 21
d = 2 → a = 24

All of these produce sequences of positive integers.

For negative d, we must ensure all 10 terms are positive by checking that a + 9d > 0.

d = −12 → a = 66 → last term = −42 (invalid)
d = −6 → a = 48 → last term = −6 (invalid)
d = −4 → a = 42 → last term = 6 (valid)

So the valid first terms are 12, 18, 21, 24, and 42.

Their sum is 12 + 18 + 21 + 24 + 42 = 117.

The answer is 117.

An arithmetic sequence of 10 positive integers has a fourth

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An arithmetic sequence of 10 positive integers has a fourth

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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