Question

4 D
The 3rd term of an arithmetic series is 12 and the 10th term is -93.
a. Find the first term and the common difference.
b .Find the term in which the series turns negative.​

Answer 1

Answer:

the 3rd term a+2d = 12.....(1)

the 10th term a+9d = ‐93.....(2)

a. (1) - (2) , we get -7d = 105

d = -15

a = 42 by putting d = -15 in (1)

b. the first negative term is -3

the term is 4th term

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Answer 2

AnswEr :

$\bigstar\:\boxed{\sf T_n = a + (n-1)d}$

☯⠀$\underline{\textsf{Third Term of AP :}}$

$:\implies\tt T_3 = 12\\\\\\:\implies\tt a + (3 - 1)d = 12\\\\\\:\implies\tt a + 2d = 12\\\\\\:\implies\tt a = 12 - 2d$

$\rule{150}{1}$

☯⠀$\underline{\textsf{Tenth Term of AP :}}$

$:\implies\tt T_{10} = -\:93\\\\\\:\implies\tt a + (10 - 1)d = - \:93\\\\\\:\implies\tt a + 9d = - \:93 \\ \\{\scriptsize\qquad\bf{\dag}\:\:\textsf{putting the value of a.}}\\\\:\implies\tt 12 - 2d + 9d = - \:93\\\\\\:\implies\tt 7d = - \:93 -12\\\\\\:\implies\tt 7d = - \:105\\\\\\:\implies\tt d = \dfrac{ - \:105}{7}\\\\\\:\implies\underline{\boxed{\red{\tt d = - \:15}}}$

☯⠀$\underline{\textsf{Using the value of d :}}$

$:\implies\tt a = 12 - 2d\\\\\\:\implies\tt a = 12 - 2( -\:15)\\\\\\:\implies\tt a = 12 + 30\\\\\\:\implies\underline{\boxed{\red{\tt a = 42}}}$

⠀

$\therefore\:\underline{\textsf{First term and CD is \textbf{42 \& - 15} respectively.}}$

$\rule{200}{2}$

☢⠀$\underline{\textsf{Arithmetic Progression :}}$

$\twoheadrightarrow\tt\: a, \: (a + d), \:(a + 2d), \:(a + 3d),...\\\\\\\twoheadrightarrow\tt\: 42,\:(42+(-15)),\:(42+2(-15)),\:(42+3(-15)), ...\\\\\\\twoheadrightarrow\tt\:42, \:(42 - 15), \:(42 - 30), \:(42 - 45),...\\\\\\\twoheadrightarrow\:\underline{\boxed{\blue{\tt42,\:27, \:12, \: -3,...}}}$

⠀

$\therefore\:\underline{\textsf{From \textbf{4th Term} series turn negative.}}$