The first term of an ap is -5 and the last term is 45. If the sum of the terms of ap is 120 then find the number of terms and the common difference
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Answer:
Let us consider an A.P. whose first term and common difference are a and d respectively.
Here, a=−5,a
n
=45,S
n
=120
Now, S
n
=
2
n
[2a+(n−1)d]=
2
n
[a+a+(n−1)d]
⇒S
n
=
2
n
[a+a
n
][a
n
=last term]
⇒120=
2
n
[−5+45]
⇒120=
2
n
×40
⇒n=
40
120×2
=6
⇒n=6
Hence, the number of terms =6
Now, a
n
=a+(n−1)d
⇒45=−5+(6−1)d
⇒45+5=5d
⇒5d=50
⇒d=10
Hence, the number of terms and the common difference of the A.P. are 6 and 10 respectively
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