Question

find the mth term of an arithmetic progression whose 12th term exceeds the 5th term by 14 and the sum of both term is 36.

Answer 1

Answer is 1+2n

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Answer 2

$\sf{\bold{\green{\underline{\underline{Given}}}}}$

• 12th term exceeds the 5th term by 14
• the sum of both terms is 36.

$\sf{\bold{\green{\underline{\underline{To\: Find}}}}}$

• The m th term of AP

$\sf{\bold{\green{\underline{\underline{Solution}}}}}$

$\sf{\large{\bold{\pink{\bigstar{\boxed{\boxed{t_n = a + ( n - 1 ) d }}}}}}}$

$\sf\implies t_5 = a + ( 5 - 1 ) d$

$\sf\implies t_5 = a + 4 d$

$\sf{\bold{\blue{\underline{\underline{t_5 = a + 4d }}}}}$

$\sf\implies t_12 = a + ( 12 - 1 ) d$

$\sf\implies t_5 = a + 11 d$

$\sf{\bold{\blue{\underline{\underline{t_{12}= a + 12d }}}}}$

• 12th term exceeds the 5th term by 14

$\sf ( a + 11d ) - ( a + 4d ) = 14$

$\sf a + 11d - a - 4d = 14$

$\sf 7d = 14$

$\sf d = \dfrac{14}{7}$

$\sf d = 2$

$\sf{\large{\bold{\orange{\bigstar{\boxed{\boxed{d = 2 }}}}}}}$

• the sum of both terms is 36

$\sf\implies ( a + 11d ) + ( a + 4d ) = 36$

$\sf\implies a + 11d + a + 4d = 36$

$\sf\implies 2a + 15d = 36$

$\sf\implies 2a + 15\times 2 = 36 ( d = 2 )$

$\sf\implies 2a + 30 = 36$

$\sf\implies 2a = 36 - 30$

$\sf\implies 2a = 6$

$\sf\implies a = \dfrac{6}{2}$

$\sf\implies a = 3$

$\sf{\large{\bold{\orange{\bigstar{\boxed{\boxed{a = 3 }}}}}}}$

__________________________

$\sf{\large{\bold{\pink{\bigstar{\boxed{\boxed{t_n = a + ( n - 1 ) d }}}}}}}$

$\sf\implies t_m = 3 + ( m - 1 )2$

$\sf\implies t_m = 3 + 2m - 2$

$\sf\implies t_m = 1 + 2m$

$\sf{\large{\bold{\orange{\bigstar{\boxed{\boxed{t_m = 1 + 2m}}}}}}}$