Question

the sum of first q terms of an ap is 63q-3q² if the p term is -60 find the value of p and also its 11th term​

Answer 1

Answer:

$\sf{The \ value \ of \ p \ is \ 21 \ and \ it's \ 11^{th} \ term}$

$\sf{is \ 0.}$

Given:

$\sf{Sum \ of \ q \ terms \ of \ an \ A.P. \ is \ 63q-3q^{2}.}$

To find:

$\sf{If \ the \ p \ term \ is \ -60 \ find \ the \ value \ of \ p}$

$\sf{and \ it's \ 11^{th} \ term.}$

Solution:

$\sf{Sum \ of \ q \ terms \ is \ 63q-3q^{2}}$

$\sf{i.e. \ S_{q}=63q-3q^{2}}$

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

$\sf{\therefore{S_{1}=63-3}}$

$\sf{\therefore{S_{1}=60}}$

$\sf{\implies{t_{1}=60}}$

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

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

$\sf{\therefore{S_{2}=114}}$

$\sf{t_{2}=S_{2}-S_{1}}$

$\sf{\therefore{t_{2}=114-60}}$

$\sf{\therefore{t_{2}=54}}$

$\sf{Now,}$

$\sf{t_{1}=a=60, \ d=t_{2}-t_{1}=54-60=-6}$

$\sf{Let, \ t_{n}=p=-60}$

$\boxed{\sf{t_{n}=a+(n-1)d}}$

$\sf{\therefore{-60=60+(n-1)\times(-6)}}$

$\sf{\therefore{-6(n-1)=-120}}$

$\sf{\therefore{n-1=\dfrac{120}{6}}}$

$\sf{\therefore{n=20+1}}$

$\sf{\therefore{n=21}}$

$\sf{Hence, \ the \ value \ of \ p \ is \ also \ 21.}$

$\sf{t_{11}=60+(11-1)\times(-6)}$

$\sf{\therefore{t_{11}=60-60}}$

$\sf{\therefore{t_{11}=0}}$

$\sf\purple{\tt{\therefore{The \ value \ of \ p \ is \ 21 \ and \ it's \ 11^{th} \ term}}}$

$\sf\purple{\tt{is \ 0.}}$