Question

Let \[ \sum_{k=1}^{n} a_k = \alpha n^2 + \beta n. \] If \( a_{10} = 59 \) and \( a_6 = 7a_1 \), then \( \alpha + \beta \) is equal to

• A. 3
• B. 5
• C. 7
• D. 12

Hint:

For $S_n = An^2+Bn$, common difference $d=2A$, first term $a=A+B$.

Answer 1

The Correct Option is B

We are given that the sum of the first \( n \) terms of a sequence is:

\[ \sum_{k=1}^{n} a_k = \alpha n^2 + \beta n \]

Given conditions:

• \( a_{10} = 59 \)
• \( a_6 = 7a_1 \)

We are required to find \( \alpha + \beta \).

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Step 1: Find the general term \( a_n \)

The general term of a sequence is obtained from:

\[ a_n = S_n - S_{n-1} \]

\[ a_n = (\alpha n^2 + \beta n) - [\alpha (n-1)^2 + \beta (n-1)] \]

\[ a_n = \alpha(2n - 1) + \beta \]

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Step 2: Use the given conditions

From \( a_{10} = 59 \):

\[ a_{10} = \alpha(19) + \beta = 59 \quad \Rightarrow \quad 19\alpha + \beta = 59 \quad (1) \]

From \( a_6 = 7a_1 \):

\[ a_6 = \alpha(11) + \beta \]

\[ a_1 = \alpha(1) + \beta \]

\[ \alpha(11) + \beta = 7(\alpha + \beta) \]

\[ 11\alpha + \beta = 7\alpha + 7\beta \]

\[ 4\alpha = 6\beta \Rightarrow 2\alpha = 3\beta \Rightarrow \beta = \frac{2}{3}\alpha \quad (2) \]

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Step 3: Solve equations (1) and (2)

\[ 19\alpha + \frac{2}{3}\alpha = 59 \]

\[ \frac{59}{3}\alpha = 59 \Rightarrow \alpha = 3 \]

\[ \beta = \frac{2}{3} \times 3 = 2 \]

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Step 4: Final result

\[ \alpha + \beta = 3 + 2 = 5 \]

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Final Answer:

\(\boxed{5}\)