If an AP has 10 terms and $a_{n}$ represents the $n^{t h}$ term of the AP, then which of the following is incorrect?

a_{1} + a_{10} = a_{2} + a_{9}

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a_{3} + a_{8} = a_{4} + a_{6}

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a_{4} + a_{7} = a_{1} + a_{10}

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a_{1} + a_{10} = a_{5} + a_{6}

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Solution

The correct option is B $a_{3} + a_{8} = a_{4} + a_{6}$
Let the first term of the AP be a and the common difference be d.
Thus, the $n^{t h}$ term, $a_{n}$ of the AP is a + (n – 1)d.
Consider: $a_{1} + a_{10} = a + a + (10 - 1) d$
$= 2 a + 9 d$
$a_{2} + a_{9} = a + (2 - 1) d + a + (9 - 1) d$
$= a + d + a + 8 d$
$= 2 a + 9 d$
$∴ a_{1} + a_{10} = a_{2} + a_{9}$
$a_{3} + a_{8} = a + (3 - 1) d + a + (8 - 1) d$
$= a + 2 d + a + 7 d$
$= 2 a + 9 d$
$a_{4} + a_{6} = a + (4 - 1) d + a + (6 - 1) d$
$= a + 3 d + a + 5 d$
$= 2 a + 8 d$
$∴ a_{3} + a_{8} \neq a_{4} + a_{6}$
$a_{4} + a_{7} = a + (4 - 1) d + a + (7 - 1) d$
$= a + 3 d + a + 6 d$
$= 2 a + 9 d$
$∴ a_{4} + a_{7} = a_{1} + a_{10}$
$a_{5} + a_{6} = a + (5 - 1) d + a + (6 - 1) d$
$= a + 4 d + a + 5 d$
$= 2 a + 9 d$
$∴ a_{1} + a_{10} = a_{5} + a_{6}$
Hence, the correct answer is option (2).

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