Question

please answer this question ​

Answer 1

Answer:

math is tuff but not impossible

Answer 2

Step-by-step explanation:

Question:-

The sum of first q terms of A.P is 63q−3q² . If its pth term is -60, Find the value of p. Also, find the 11th term of this AP

Solution:-

$\sf The\: sum \: of \: first \: q \: terms = 63q - 3q^{2}$

$\sf \longrightarrow S_{q}= 63q - 3q^{2}$

$\sf \longrightarrow S_{1}= 63(1) - 3(1)^{2}$

$\sf \longrightarrow S_{1}= 60 = a_{1}$

$\sf Substituting \: 2 \: in \: the \: place \: of \: q$

$\sf \longrightarrow S_{2}= 63(2) - 3(2)^{2}$

$\sf \longrightarrow S_{2}= 126 - 12$

$\sf \longrightarrow S_{2}= 114 = a_{1} + a_{2}$

$\sf \longrightarrow a_{2} = S_{2} - S_{1}$

$\sf \longrightarrow a_{2} = 114 - 60 = 54$

Common difference = Second term - first term

$\sf \longrightarrow d = a_{2} - a_{1}$

$\sf \longrightarrow d = 54 - 60 = -6$

$\sf Given, \: pth \: term = -60$

$\sf \longrightarrow a_{p} = -60$

$\sf \longrightarrow a + (p - 1)d = -60$

$\sf \longrightarrow 60 + (p - 1)(-6)= -60$

$\sf \longrightarrow (p - 1)(-6) = -60 -60$

$\sf \longrightarrow p - 1 = \dfrac{-120}{-6}$

$\sf \longrightarrow p - 1 = 20$

$\sf \longrightarrow p= 20 + 1 = 21$

$\underline{\boxed{\sf \therefore The \: value \: of \: p = 21}}$

$\sf Now, \: let \: us \: find \: the \: 11th\: term$

$\sf a_{11} = a + 10d$

$\sf = 60 + 10(-60)$

$\sf = 60 - 60$

$\sf = 0$

$\underline{\boxed{\sf \therefore The \: 11th\: term = 0}}$