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Question

How many terms of the series \(1+3+5+7+...\) amount to a sum equal to \(12345678987654321\)?

This question was previously asked in
NDA 2 2025 GAT Question Paper (14-Sep-2025)
The correct answer is
111111111

Understanding the Arithmetic Series

The given series is \(1+3+5+7+...\). This is an arithmetic series where:

  • The first term (\(a\)) is \(1\).
  • The common difference (\(d\)) is \(3 - 1 = 2\).

We need to find the number of terms (\(n\)) such that the sum of these terms (\(S_n\)) equals \(12345678987654321\).

Calculating the Sum of the Series

The formula for the sum of the first \(n\) terms of an arithmetic series is:

\(S_n = \frac{n}{2} [2a + (n-1)d]\)

Substitute the values \(a=1\) and \(d=2\) into the formula:

\(S_n = \frac{n}{2} [2(1) + (n-1)2]\)

\(S_n = \frac{n}{2} [2 + 2n - 2]\)

\(S_n = \frac{n}{2} [2n]\)

\(S_n = n^2\)

This shows that the sum of the first \(n\) odd numbers is equal to the square of \(n\). In this specific series, the sum of the first \(n\) terms is \(n^2\).

Finding the Number of Terms

We are given that the sum \(S_n = 12345678987654321\). Therefore, we have the equation:

\(n^2 = 12345678987654321\)

To find \(n\), we need to calculate the square root of \(12345678987654321\).

Let's examine the pattern of squares of numbers consisting of ones:

Number Square Pattern
\(1\) \(1^2 = 1\) \(1\)
\(11\) \(11^2 = 121\) \(121\)
\(111\) \(111^2 = 12321\) \(12321\)
\(1111\) \(1111^2 = 1234321\) \(1234321\)
\(111111111\) \(111111111^2 = 12345678987654321\) \(12345678987654321\)

By observing the pattern, we can see that the square root of \(12345678987654321\) is \(111111111\).

So, \(n = \sqrt{12345678987654321} = 111111111\)

Conclusion

The number of terms of the series \(1+3+5+7+...\) that amount to a sum equal to \(12345678987654321\) is \(111111111\).

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Similar Questions

  1. The fifth term of an AP of n terms, whose sum is n 2– 2n, is

  2. A person is to count 4500 notes. Let a ndenote the number of notes he counts in the nth minute. If a 1= a 2= a 3= … = a 10 = 150, and a 10 , a 11 , a 12 , … are in AP with the common difference -2, then the time taken by him to count all the notes is

  3. If \({S_n} = nP + \frac{{n\left( {n - 1} \right)Q}}{2}\) , where S ndenotes the sum of the first n terms of an AP, then the common difference is

  4. If the ratio of AM to GM of two positive numbers a and b is 5 : 3 then a : b is equal to

  5. Let x, y, z be positive real numbers such that x, y, z are in GP and tan -1 x, tan -1 y and tan -1 z are in AP. Then which one of the following is correct?

  6. If x 1and x 2are positive quantities, then the condition for the difference between the arithmetic mean and the geometric mean to be greater than 1 is

  7. If y = x + x 2+ x 3+ … up to infinite terms where x < 1, then which one of the following is correct?

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Important Questions from Sequences and Series

  1. The fifth term of an AP of n terms, whose sum is n 2– 2n, is

  2. A person is to count 4500 notes. Let a ndenote the number of notes he counts in the nth minute. If a 1= a 2= a 3= … = a 10 = 150, and a 10 , a 11 , a 12 , … are in AP with the common difference -2, then the time taken by him to count all the notes is

  3. \(\mathop {\lim }\limits_{n \to \infty } {\left( {1 - \frac{1}{{2n}}} \right)^{n + 1}}\) is equal to
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  5. What is the sum of the first 12 terms of an arithmetic progression if the 3rd term is -13 and the 6th term is -4?

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