Question ID: #884
Let $\alpha_{1}, \alpha_{2}, \alpha_{3}, \alpha_{4}$ be an A.P. of four terms such that each term of the A.P. and its common difference $d$ are integers. If $\alpha_{1}+\alpha_{2}+\alpha_{3}+\alpha_{4}=48$ and $\alpha_{1}\alpha_{2}\alpha_{3}\alpha_{4}+d^{4}=361$, then the largest term of the A.P. is equal to
- (1) 27
- (2) 24
- (3) 21
- (4) 23
Solution:
Let the terms be $a-3d, a-d, a+d, a+3d$.
Common difference of this A.P. is $2d$. Let original common difference be $D$, so $D = 2d$.
Sum $= 4a = 48 \Rightarrow a = 12$.
Product $+ D^4 = 361$ (Wait, the question says “common difference $l$ are integers… + $l^4$”. Let’s assume the variable in the term is the common difference $l$).
Let terms be $a-3\delta, a-\delta, a+\delta, a+3\delta$ where common difference $l = 2\delta$.
$(a-3\delta)(a-\delta)(a+\delta)(a+3\delta) + (2\delta)^4 = 361$
$(a^2 – 9\delta^2)(a^2 – \delta^2) + 16\delta^4 = 361$
Substitute $a=12$:
$(144 – 9\delta^2)(144 – \delta^2) + 16\delta^4 = 361$
$20736 – 144\delta^2 – 1296\delta^2 + 9\delta^4 + 16\delta^4 = 361$
$25\delta^4 – 1440\delta^2 + 20375 = 0$
Divide by 25:
$\delta^4 – \frac{1440}{25}\delta^2 + \frac{20375}{25} = 0$
$\delta^4 – 57.6\delta^2 + 815 = 0$
(Re-evaluating based on provided solution snippet which uses factorization):
$(5\delta^2 – 144)^2 = 19^2$
$5\delta^2 – 144 = 19 \Rightarrow 5\delta^2 = 163 \Rightarrow \delta^2 = 32.6$ (Not integer solution for $l$)
$5\delta^2 – 144 = -19 \Rightarrow 5\delta^2 = 125 \Rightarrow \delta^2 = 25 \Rightarrow \delta = 5$
Common difference $l = 2\delta = 10$.
The terms are $12-15, 12-5, 12+5, 12+15 \Rightarrow -3, 7, 17, 27$.
Largest term is 27.
Ans. (1)
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