Question

\( t_1, t_2, t_3, ..., t_n \) are positive integers, \( S_n = t_1 + t_2 + t_3 + ... + t_n \), \( S_1 = 1^2 \), \( S_2 = 3^2 \), \( S_3 = 6^2 \), \( S_4 = 10^2 \), \( S_5 = 15^2 \) and similarly other terms are there. Following this pattern, if \( S_{10} = k^2 \) then \( k = \)

• A. 55
• B. 45
• C. 36
• D. 21

Hint:

Recognizing common number sequences is a key skill. The sequence 1, 3, 6, 10, 15, ... represents triangular numbers. Knowing the formula \( T_n = n(n+1)/2 \) makes solving such problems much faster.

Answer 1

The Correct Option is A

We are given a pattern for \( S_n \). Let's analyze the base of the square term in each \( S_n \).
Let \( S_n = (B_n)^2 \).
For \( n=1 \), \( S_1 = 1^2 \implies B_1 = 1 \).
For \( n=2 \), \( S_2 = 3^2 \implies B_2 = 3 \).
For \( n=3 \), \( S_3 = 6^2 \implies B_3 = 6 \).
For \( n=4 \), \( S_4 = 10^2 \implies B_4 = 10 \).
For \( n=5 \), \( S_5 = 15^2 \implies B_5 = 15 \).
The sequence of the bases \( B_n \) is \( 1, 3, 6, 10, 15, ... \).
Let's find the relationship between consecutive terms:
\( B_1 = 1 \)
\( B_2 = 3 = 1 + 2 \)
\( B_3 = 6 = 1 + 2 + 3 \)
\( B_4 = 10 = 1 + 2 + 3 + 4 \)
\( B_n \) is the sum of the first \( n \) natural numbers, also known as the \(n\)-th triangular number.
The formula for the \(n\)-th triangular number is \( B_n = \frac{n(n+1)}{2} \).
We need to find \( k \) such that \( S_{10} = k^2 \). This means \( k = B_{10} \).
Using the formula for \( n=10 \):
\( k = B_{10} = \frac{10(10+1)}{2} = \frac{10 \times 11}{2} = \frac{110}{2} = 55 \).
Thus, the value of \( k \) is 55.