The sum of first $q$ terms of an A.P. is $63q – 3q^2$. If its $p$th term is $-60$, find the value of $p$, Also, find the 11th term of this A.P.

Given:

The sum of first $q$ terms of an A.P. is $63q – 3q^2$ and its $p$th term is $-60$.

To do:

We have to find the value of $p$ and $11^{th}$ term of the given A.P.

Solution:

$S_{q} =63q-3q^{2}$

For $q=1,\ S_{1} =63\times 1-3\times 1^{2}=63-3=60$

Therefore, first term $a=60$

For $n=2,\ S_{2} =63\times 2 - 3\times 2^2=126-12=114$

$\therefore$ Second term of the A.P.$=S_{2} -S_{1}$

$=114-60$

$=54$

Common difference of the A.P., $d=$second term $-$ first term

$=54-60=-6$

We know that,

$a_{q}=a+(q-1)d$

$a_{q}=60+( q-1) \times (-6)$

$-60=60-6q+6$

$6q=66+60$

$6q=126$

$q=21$

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

$=60+10\times (-6)$

$=60-60$

$=0$

Therefore, the value of $q$ is $21$ and the $11^{th}$ term of the given A.P. is $0$. 

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Updated on: 10-Oct-2022

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