Mathematics · Quantitative Aptitude

Number Sums and Series

287 Questions

Number sums and series questions test the ability to find patterns and calculate the sum of number sequences. These problems are a staple in the quantitative aptitude sections of various competitive exams. Practice this collection to improve speed and accuracy in solving arithmetic and geometric series problems.

Sum of consecutive integersGeometric series sumsRatio and proportion sumsOdd and even number propertiesRoman numeral calculations

Number Sums and Series Questions

Multiple choice

Find three numbers in G.P whose sum is 19 and product is 216

  1. 4, 6, 9

  2. 3, 7, 9

  3. 11, 6, 2

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For three numbers in G.P, let them be a/r, a, ar. Sum = a/r + a + ar = 19. Product = (a/r) x a x ar = a³ = 216, so a = 6. Substituting: 6/r + 6 + 6r = 19. Divide by 6: 1/r + 1 + r = 19/6. Multiply by r: 1 + r + r² = 19r/6. This gives r = 1.5 or 2/3. So numbers are 4, 6, 9 (with r = 1.5) or 9, 6, 4 (with r = 2/3). Option A matches one valid ordering.

Multiple choice

The sum of the infinite series 1 + 2/3 + 4/9 +......... is

  1. 1/3

  2. 3

  3. 2/3

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

This is an infinite geometric series with first term a = 1 and common ratio r = 2/3. The sum of an infinite GP is S = a/(1-r) when |r| < 1. Here, S = 1/(1 - 2/3) = 1/(1/3) = 3. The series converges because 2/3 < 1.

Multiple choice

The sum of three numbers in G.P is 70. If the two extrmes be multiplied each by 4 and the mean by 5, the products are in AP. The numbers are

  1. 12, 18, 40

  2. 10, 20, 40

  3. 40, 20, 10

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the three numbers in G.P. be a/r, a, ar. Their sum is 70. When extremes are multiplied by 4 and mean by 5, we get 4a/r, 5a, 4ar. For these to be in A.P., the middle term must equal the average of first and third: 5a = (4a/r + 4ar)/2. Solving with sum=70 gives the numbers 10, 20, 40. Check: 10+20+40=70 and 4×10, 5×20, 4×40 = 40, 100, 160 which is an A.P. with common difference 60.

Multiple choice

Three numbers are in A.P and their sum is 21. If 1 , 5 , 15 are added to them respectively, they form a G.P. The numbers are

  1. 5, 7, 9

  2. 9, 5, 7

  3. 7, 5, 9

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the AP be (a-d), a, (a+d). Sum = 3a = 21, so a = 7. After adding 1, 5, 15 respectively: (8-d), 12, (22+d). For GP: 12^2 = (8-d)(22+d). Solving d^2 + 14d - 32 = 0 gives d = 2 or d = -16. For d = 2: numbers are 5, 7, 9. For d = -16: numbers are 23, 7, -9. Both work, but only 5, 7, 9 appears in the options.

Multiple choice

The sum of four numbers in G.P is 60 and the A.M of the 1st and the last is 18. The numbers are

  1. 4, 8, 16, 32

  2. 4, 16, 8, 32

  3. 16, 8, 4, 20

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the GP be a, ar, ar^2, ar^3. Given a(1+r+r^2+r^3) = 60 and (a+ar^3)/2 = 18, so a+ar^3 = 36. Dividing the sum equation by this gives (1+r+r^2+r^3)/(1+r^3) = 5/3. This leads to 2r^3 - 3r^2 - 3r + 2 = 0, which has r = 2 as a solution. With r = 2, a(1+8) = 36 gives a = 4. The numbers are 4, 8, 16, 32.

Multiple choice

The sum of all odd numbers between 200 and 300 is

  1. 11600

  2. 12490

  3. 12500

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The odd numbers between 200 and 300 are 201, 203, 205, ..., 299. This is an arithmetic progression with first term 201, last term 299, and 50 terms. Sum = n/2 × (first + last) = 50/2 × (201 + 299) = 25 × 500 = 12500.