Question

The A.P. in which 4^th term is –15 and 9^th term is –30. Find the sum of the first 10 numbers.

Answer 1

It is given that,
t₄ = –15
t₉ = –30

Now,
$$\begin{gathered} t_n=a+\left(n-1\right)d \\ {t}_4=a+\left(4-1\right)d \\ \Rightarrow -15=a+3d \\ \Rightarrow a=-15-3d...\left(1\right) \\ {t}_9=a+\left(9-1\right)d \\ \Rightarrow -30=a+8d \\ \Rightarrow a+8d=-30 \\ \Rightarrow -15-3d+8d=-30\left(\mathrm{from}\left(1\right)\right) \\ \Rightarrow -15+5d=-30 \\ \Rightarrow 5d=-30+15 \\ \Rightarrow 5d=-15 \\ \Rightarrow d=-3 \\ \Rightarrow a=-15-3\left(-3\right)\left(\mathrm{from}\left(1\right)\right) \\ \Rightarrow a=-15+9 \\ \Rightarrow a=-6 \end{gathered}$$

Hence, the given A.P. is –6, –9, –12, ....

Now,
$$\begin{gathered} S_n=\frac{n}{2}\left(2a+\left(n-1\right)d\right) \\ {S}_{10}=\frac{10}{2}\left(2a+\left(10-1\right)d\right) \\ =5\left(2\left(-6\right)+9\left(-3\right)\right) \\ =5\left(-12-27\right) \\ =5\left(-39\right) \\ =-195 \end{gathered}$$

Hence, the sum of the first 10 numbers is –195.