Find the number of terms of an A.P. whose 1st term is 5, the common difference is 5 and
the last term is 45.
Answers
an =a(n-1)d
45=5+(n-1)5
45-5=5n-5
40+5=5n
45\5=n
n=9
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The first term of an A.P. is 5 and last term is 45. If the sum of the terms of the A.P. 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 common difference and the number of terms in A.P. are 10 and 6 respectively.