How many terms are there in the AP whose first term and 6 th term are - 8 and 12 respectively and the sum of all its terms is 120?
Please answer fast I have to submit at 10:30
Answers
Step-by-step explanation:
Answer:
S_{120}=27120S
120
=27120
Step-by-step explanation:
\begin{gathered}Let \: a \: and \: d \: are \\first\:term \:and \: common\: difference\:of \: an \:A.P\end{gathered}
Letaanddare
firsttermandcommondifferenceofanA.P
\begin{gathered}We \: know \: that ,\\\boxed {n^{th}\: term =a_{n}=a+(n-1)d}\end{gathered}
Weknowthat,
n
th
term=a
n
=a+(n−1)d
According to the problem given,
First\:term =a= -12Firstterm=a=−12
6^{th}\:term = 86
th
term=8
\implies a+5d = 8⟹a+5d=8
\implies -12+5d = 8⟹−12+5d=8
\implies 5d = 8+12⟹5d=8+12
\implies 5d = 20⟹5d=20
\implies d = \frac{20}{5}⟹d=
5
20
\implies d = 4⟹d=4
\begin{gathered}Now, \\Sum \:of \: n \:terms (S_{n})=\frac{n}{2}[2a+(n-1)d]\end{gathered}
Now,
Sumofnterms(S
n
)=
2
n
[2a+(n−1)d]
Here, a = -12 , \:d=4,\:n=120Here,a=−12,d=4,n=120
S_{120}=\frac{120}{2}[2\times (-12)+(120-1)4]S
120
=
2
120
[2×(−12)+(120−1)4]
=60(-24+119\times 4)=60(−24+119×4)
=60(-24+476)=60(−24+476)
=60\times 452=60×452
=27120=27120
Therefore,
S_{120}=27120S
120
=27120
•••♪
Answer:
a+(6-1)d=12
-8+5d=12
5d=20
d=4
n/2(2a+(n-1)d)=120