To solve this problem, we begin by analyzing the formula for the sum of the first \( n \) terms of an Arithmetic Progression (A.P.). Given that the sum, \( S_n \), is \( 3n^2 + 5n \), we aim to determine the first term and the common difference of the A.P.
The formula for the sum of the first \( n \) terms of an A.P. is:
\(S_n = \frac{n}{2} \left(2a + (n-1)d\right)\)
For the given A.P., we equate this standard formula to the expression \( 3n^2 + 5n \):
\(\frac{n}{2} \left(2a + (n-1)d\right) = 3n^2 + 5n\)
Simplifying, we get:
\(2a + (n-1)d = 6n + 10\)
To find \( a \) and \( d \), we'll use specific values of \( n \) (e.g., \( n=1 \) and \( n=2 \)):
- For \( n = 1 \): \( 2a = 6 \cdot 1 + 10 = 16 \) gives \( a = 8 \).
- For \( n = 2 \): \( 2a + d = 6 \cdot 2 + 10 = 22 \).
Substituting \( 2a = 16 \) into \( 2a + d = 22 \), we find:
\(16 + d = 22 \Rightarrow d = 6\)
Thus, the A.P. has the first term \( a = 8 \) and common difference \( d = 6 \).
Next, we calculate the sum of squares of the first 10 terms of the A.P.
The first 10 terms of the A.P. are: \(a, a+d, a+2d, \ldots\). Specifically, these terms are \(8, 14, 20, \ldots\).
The formula for the sum of squares of these terms is:
\(\Sigma_{i=1}^{10} \left(a_i^2\right)\)
where \(a_i = a + (i-1) \cdot d\).
We calculate:
- \((a_1)^2 = 8^2 = 64\)
- \((a_2)^2 = 14^2 = 196\)
- \((a_3)^2 = 20^2 = 400\)
- Continue this for all 10 terms: \(64, 196, 400, 676, 1024, 1444, 1936, 2500, 3136, 3844.\)
The sum is:
\(64 + 196 + 400 + 676 + 1024 + 1444 + 1936 + 2500 + 3136 + 3844 = 15220\)
Thus, the sum of squares of the first 10 terms of the A.P. is 15220. Hence, the correct answer is:
15220
Similar Questions
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Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.
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