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Question
In an A.P. the first term is – 5 and the last term is 45. If the sum of all numbers in the A.P. is 120, then how many terms are there? What is the common difference?
Solution
It is given that,
a = – 5
l = 45
Sn = 120
Now,
\[S_n = \frac{n}{2}\left( a + l \right)\]
\[ \Rightarrow 120 = \frac{n}{2}\left( - 5 + 45 \right)\]
\[ \Rightarrow 120 = \frac{n}{2}\left( 40 \right)\]
\[ \Rightarrow 120 \times 2 = n\left( 40 \right)\]
\[ \Rightarrow 240 = n\left( 40 \right)\]
\[ \Rightarrow n = \frac{240}{40}\]
\[ \Rightarrow n = 6\]
Hence, there are 6 terms.
Also,
\[l = a + \left( 6 - 1 \right)d\]
\[ \Rightarrow 45 = - 5 + 5d\]
\[ \Rightarrow 45 + 5 = 5d\]
\[ \Rightarrow 5d = 50\]
\[ \Rightarrow d = 10\]
Hence, the common difference is 10.
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