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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?
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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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