If t5,t10 and t25 are 5th,10th and 25th terms of an AP respectively, then the value of ∣∣
∣∣t5t10t2551025111∣∣
∣∣ is
A
−40
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B
1
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C
−1
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D
0
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E
40
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Solution
The correct option is B0 Let a and d be the first and common difference of an AP. Then, t5=a+4d t10=a+9d and t25=a+24d Now, Let △=∣∣
∣∣t5t10t2551025111∣∣
∣∣ =∣∣
∣∣a+4da+9da+24d51025111∣∣
∣∣ On applying operation C2→C2−C1 and C3→C3−C1, we get △=∣∣
∣∣a+4d5d20d5520100∣∣
∣∣ On expanding along R3, we get △=100d−100d=0.