Question

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

Answer 1

Given :-

• The sum of first q terms of an A.P is $\sf63q - 3q^{2}$
• Its pth term is -60.

To Find :-

• Value of P and 11th term of this A.P.

Solution :-

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

$\sf \pink{ \: \: \: \: \: \: \: \: \: \:\::\implies S_{q}= 63q - 3q^{2}}$

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

$\sf \: \: \: \: \: \: \: \: \: \pink{\:\::\implies S_{1}= 60 = a_{1}}$

$\sf \underline {Substituting \: 2 \: in \: the \: place \: of \: q}$

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

$\sf \: \: \: \: \: \: \: \: \: \:\::\implies S_{2}= 126 - 12$

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

$\sf \: \: \: \: \: \: \: \: \: \:\::\implies a_{2} = S_{2} - S_{1}$

$\sf \: \: \: \: \: \: \: \: \:\pink{ \:\::\implies a_{2} = 114 - 60 = 54}$

• Common difference = Second term - first term

$\sf \: \: \: \: \: \: \: \: \: \:\::\implies d = a_{2} - a_{1}$

$\sf \: \: \: \: \: \: \: \: \: \:\::\implies d = 54 - 60 = -6$

$\sf We \: are \: given:- \green{ \: pth \: term = -60}$

$\sf \: \: \: \: \: \: \: \: \:\pink{ \:\::\implies a_{p} = -60}$

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

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

$\sf \: \: \: \: \: \: \: \: \: \:\::\implies (p - 1)(-6) = -60 -60$

$\sf \: \: \: \: \: \: \: \: \: \:\::\implies p - 1 = \cancel{\dfrac{-120}{-6}}$

$\sf \: \: \: \: \: \: \: \: \: \:\::\implies p - 1 = 20$

$\sf \: \: \: \: \: \: \: \: \: \pink{\:\::\implies p= 21}$

$\therefore\:\underline{\textsf{ Value of p is \textbf{21}}}.$

$\sf \underline{ 11th\: term}:-$

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

$\sf \: \: \: \: \: \: \: \: \: \:\::\implies a_{11}= 60 + 10(-60)$

$\sf \: \: \: \: \: \: \: \: \: \:\::\implies a_{11}= 60 - 60$

$\sf \red{\: \: \: \: \: \: \: \: \: \:\::\implies a_{11}= 0}$

$\therefore\:\underline{\textsf{ The 11th term is \textbf{0}}}.$

Answer 2

Step-by-step explanation:

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Hope the answer is helpful to you

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