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?

Answer 1

It is given that,

a = – 5

l = 45

S_n = 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.