Which term of the AP: $ -2,-7,-12, \ldots $ will be $ -77 $ ? Find the sum of this AP upto the term $ -77 $.
Given:
Given A.P. is $-2, -7, -12, …$
To do:
We have to find which term of the given A.P. is $-77$ and the sum of this A.P. upto the term $-77$.
Solution:
Here,
$a_1=-2, a_2=-7, a_3=-12$
Common difference $d=a_2-a_1=-7-(-2)=-7+2=-5$
We know that,
nth term $a_n=a+(n-1)d$
Therefore,
$a_{n}=-2+(n-1)(-5)$
$-77=-2+n(-5)-1(-5)$
$-77+2=-5n+5$
$75+5=5n$
$5n=80$
$n=\frac{80}{5}$
$n=16$
We know that,
Sum of $n$ terms in an A.P. $S_n = \frac{n}{2}[2a + (n – 1) d]$
Therefore, the sum of the given A.P. upto the term $-77$ is,
$S_{16} = \frac{16}{2}[2 (-2) + (16 – 1)(-5)]$
$= 8[-4 + (-75)]$
$= 8(-79)$
$=-632$
Therefore, $-77$ is the 16th term of the given A.P. and the sum of the A.P. upto the term $-77$ is $-632$.
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