The ninth term of an A.P is equal to seven times the second term and twelfth term exceeds five times the third term by 2. Find the first term and the common difference. [3 MARKS]

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Solution

Formula: 1 Mark
Value of 'a' and 'd': 1 Mark each

Let the first term of A.P be 'a' and common difference be d.

Given, the ninth term of an A.P is equal to seven times the second term.
$\Rightarrow a_{9} = 7 a_{2}$
$\Rightarrow a + 8 d = 7 (a + d)$ $[∵ T_{n} = a + (n - 1) d]$
$\Rightarrow a + 8 d = 7 a + 7 d$
$\Rightarrow 6 a - d = 0$
$\Rightarrow 6 a = d . . . (1)$

Given, twelfth term exceeds five times the third term by 2
$\Rightarrow a_{12} = 5 a_{3} + 2$
$\Rightarrow a + 11 d = 5 (a + 2 d) + 2$ $[∵ T_{n} = a + (n - 1) d]$
$\Rightarrow a + 11 d = 5 a + 10 d + 2$
$\Rightarrow 4 a - d = - 2... (2)$

Substituting (1) in (2)
$4 a - d = - 2$
$\Rightarrow 4 a - (6 a) = - 2$
$\Rightarrow - 2 a = - 2$
$\Rightarrow a = 1$

From (1),
$6 a = d$
$\Rightarrow d = 6 \times (1) = 6$

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