Question

the 7th term of an A.P. is -39/12 and the 15th term is -103/12 what is the 27th term​

Answer 1

ANSWER-

$\sf \bold{Given\::}$

$\sf \bullet \: T_7 = \dfrac{-39}{\:\: 12} \\\\\\\sf \bullet \: T_{15} = \dfrac{-103}{12}$

$\sf \bold{To\:find\::}$

$\sf 27^{th} \: term \: = T_{27}$

$\sf \bold{Formula\:used\::}$

$\sf T_n = a\:+\:(n-1)d\\\\\sf Here,\\\\\sf \bullet \: a \: =\: first \: term\\\sf \bullet \: d\: =\: common \: difference\\\sf \bullet \: T_n = n^{th} term$

$\sf \bold{Solution\::}$

$\sf \bullet T_7 = a + (7-1)d\\\\\sf \implies \dfrac{-39}{12}\: =\: a + 6d.......... equation 1\\\\\\\sf \bullet \: T_{15} = \: a + (15-1)d\\\\ \sf \implies \dfrac{-103}{12} = a + 14d ........ equation 2$

$\sf On \: subracting \: equation\: 1 \:from \:equation\: 2\: we \:get;\\\\\sf \implies \dfrac{-103}{12} - \: \dfrac{-39}{12} \: = (a\: + 14d) - (a+6d)\\\\\\\sf \implies \dfrac{-103 \:-\:(-39)}{12} = \: 8d\\\\\\\sf \implies \: d = \dfrac{-64}{12 \times 8} \\\\\\\sf \implies \: d = \dfrac{-8}{12}$

Putting value of d in equation 1 , we get;

$\sf \implies \dfrac{-39}{12} = a + 6 \times \dfrac{-8}{12}\\\\\\\sf \implies a = \dfrac{-39}{12} - \dfrac{-48}{12}\\\\\\\sf \implies a = \dfrac{-39\:+48}{12}\\\\\\\sf \implies \: a = \dfrac{9}{12}$

Now,

$\sf \bullet \: a = \dfrac{9}{12}\\\\\sf \bullet \: d = \dfrac{-8}{12}\\\\\sf \bullet \: n = 27\\\\\\\sf \implies \: T_{27} = a \:+(27-1)d\\\\\sf \implies \: T_{27} = \dfrac{9}{12} \: + \: 26 \times \dfrac{(-8)}{12} \\\\\\\sf \implies \: T_{27} = \dfrac{9\:- 208}{12}\\\\\\\sf \implies \: T_{27} = \dfrac{-199}{12}$

$\sf \bold{ANSWER\::} \:\:\: 27^{th} \:term\:of \: given \: AP = \bold{\dfrac{-199}{12}}$