Question

The sum of first q terms of an AP is ( 63q - 3q^2 ). If its pth term is -60, find the value of p. Also, find the 11th term of its AP.

Answer 1

$\underline{\underline{\large{\mathfrak{Solution : }}}}$



$\underline{\textsf{Given : }} \\ \\ \mathsf{\implies S_{q} \: = \: 63q \: - \: 3q^{2}} \\ \\ \mathsf{\implies T_{p} \: = \: -60} \\ \\ \underline{\textsf{Let,} }\\ \\ \mathsf{\implies First \: term \: = \: a } \\ \\ \mathsf{\implies Common \: difference \: = \: d } \\ \\ \underline{\textsf{To Find : }} \\ \\ \mathsf{\implies p \: = \: ?} \\ \\ \mathsf{\implies T_{11} \: = \: ?}$



$\underline{\textsf{We know that : } }\\ \\ \mathsf{\implies S_{n} \: - \: S_{n \: - \: 1} \: = \: T_{n}}$



$\underline{\textsf{Now : }} \\ \\ \mathsf{\implies S_{q} \: - \: S_{q \: - \: 1} \: = \: T_{q}} \\ \\ \mathsf{\implies ( 63q \: - \: 3q^{2}) \: - \: \{ 63(q \: - \: 1) \: - \: 3(q \: - \: 1)^{2} \} \: = \: T_{q} } \\ \\ \mathsf{ \implies63q \: - \: 3 {q}^{2} \: - \: \{63q \: - \: 63 \: - \: 3( {q}^{2} \: + \: 1 \: - \: 2q) \} \: = \: T_{q}} \\ \\ \mathsf{ \implies63q \: - \: 3 {q}^{2} \: - \: \{ 63q \: - \: 63 \: - \: 3 {q}^{2} \: - \: 3 \: + \: 6q \} \: = \: T_{q}} \\ \\ \mathsf{ \implies63q \: - \: 3 {q}^{2} \: - \: \{ 63q \: - \: 66\: - \: 3 {q}^{2} + \: 6q \} \: = \: T_{q}} \\ \\ \mathsf{ \implies \cancel{63q} \: - \: \cancel{3 {q}^{2}} \: - \: \cancel{ 63q} \: + \: 66\: + \: \cancel{3 {q}^{2} } \: - \: 6q \: = \: T_{q}} \\ \\ \mathsf{ \therefore \quad \: T_{q} \: = \: 66 \: - \: 6q}$



$\underline{\textsf{Using Formula : }} \\ \\ \boxed{\mathsf{\implies T_{n} \: = \: a \: + \: ( n \: - \: 1)d}}$



$\mathsf{\implies T_{q} \: = \: a \: + \: (q \: - \: 1)d} \\ \\ \mathsf{\implies 66 \: - \: 6q \: = \: a \: + \: ( q \: - \: 1)d } \\ \\ \textsf{When , q = 1 : } \\ \\ \mathsf{\implies 66 \: - \: 6 \: \times \: 1 \: = \: a \: + \: ( 1 \: - \: 1)d } \\ \\ \mathsf{\implies 66 \: - \: 6 \: = \: a \: + \: 0 \: \times \: d } \\ \\ \mathsf{\implies 60 \: = \: a } \\ \\ \mathsf{\therefore \: \quad a \: = \: 60 \qquad...(1)}$



$\textsf{When , q = 2 : } \\ \\ \mathsf{\implies 66 \: - \: 6 \: \times \:2 \: = \: a \: + \: ( 2 \: - \: 1)d } \\ \\ \mathsf{\implies 66 \: - \: 12 \: = \: a \: + \: 1\: \times \: d } \\ \\ \mathsf{\implies 54 \: = \: a \: + \: d} \\ \\ \mathsf{\therefore \quad \: a \: + \: d \: = \: 54 \qquad...(2)}$




$\textsf{Subtract eq.(1) from eq.(2) : } \\ \\ \mathsf{\implies \cancel{a }\: + \: d \: - \: \cancel{a} \: = \: 54 \: - \: 60} \\ \\ \mathsf{\therefore \quad \: d \: = \: -6}$



$\underline{\mathsf{To \: Find \longrightarrow Value \: of \: p \: : }} \\ \\ \\ \underline{\textsf{Using Formula : }} \\ \\ \boxed{\mathsf{\implies T_{n} \: = \: a \: + \: ( n \: - \: 1)d}}$



$\mathsf{\implies T_{p} \: = \: 60 \: + \: (p \: - \: 1)(-6)} \\ \\ \mathsf{\implies -60 \: = \: 60 \: + \: ( p \: - \: 1)(-6)} \\ \\ \mathsf{\implies -60 \: - \: 60 \: = \: (p \: - \: 1)(-6)} \\ \\ \mathsf{\implies -120 \: = \: (p \: - \: 1)(-6)} \\ \\ \mathsf{\implies \dfrac{-120}{-6} \: = \: (p \: - \: 1)} \\ \\ \mathsf{\implies 20 \: = \: p \: - \: 1} \\ \\ \mathsf{\implies 20 \: + \: 1 \: = \: p } \\ \\ \mathsf{\therefore \quad \: p \: = \: 21}$




$\underline{\mathsf{To \: Find \longrightarrow 11th \: term : }}$

$\mathsf{\implies T_{n} \: = \: a \: + \: (n \: - \: 1)d} \\ \\ \mathsf{\implies T_{11} \: = \:60 \: + \: ( 11 \: - \: 1)( - 6)} \\ \\ \mathsf{\implies T_{11} \: = \: 60 \: + \: 10(-6)} \\ \\ \mathsf{\implies T_{11} \: = \: 60 \: - \: 60 } \\ \\ \mathsf{\therefore \quad \: T_{11} \: = \: 0}$




$\boxed{\mathsf{Hence \: , \: p \: = \: 21 \: and \: T_{11} \: = \: 0 .}}$

Answer 2

HEY THERE!!

Question:-

The sum of first q terms of an AP is ( 63q - 3q2 ). If its pth term is -60, find the value of p. Also, find the 11th term of its Arithmetic Progression.

Method of Solution;-

Let to be first term of Arithmetic Sequence or Progression= "a'

And,

Let to be common Difference of Arithmetic Sequence or Progression="d"

Let Sq denotes the sum of the first qth term of the Arithmetic Sequence or Progression.

Thus,

Sq=63q-3q² (Given)

•°• Sq-1 =63(q-1)-3(q-1)²

            = 63q-63-3(q²-2q+1)

          = -3q²+69q-66

According to the formula of Tn =Sn - Sn-1

Here's, Tn is Equal to Tq.

Tq=Sq-Sq-1

  = (63q-3q²)-(-3q²+69q-66)

 = 63q-3q²+3q²-69q+66

 = -6q+66

Now, According to the Question as per as follows of terms;-

Tp=-60 (Given)

From Equation (1)

Tp= -6p+66

-60=-6p+66

-60-66=-6p

-126=-6p

126=6p

•°• P=126/6

      = 21

Hence, Pth term of Arithmetic Sequence or Progression= 21

Again, According to the Question statement;

==================================

*the 11th term of its Arithmetic Sequence or Progression*

Substitute the value of q in Equation (-6q+66=-60)

Tq=-6q+66

 T11 => -6(11)+66

T11=> -66+66

 T11=> 0

Hence, Value of T11= 0

Conclusion :-

Value of P of this Arithmetic Sequence or Progression = 21

11th term of its Arithmetic Sequence or Progression= 0

Thanks!